JC2 H2 Maths Bukit Timah | Trig General Solutions

Did you know? A calculator may give one valid solution to sin(2x)=1/2 while the question actually has four solutions on the interval 0≤x<2π. That is not necessarily a calculator error. It is a difference between a principal inverse-trigonometric value and the complete periodic solution set. One missing branch or one missing cycle is enough to lose otherwise straightforward marks.

This JC2 H2 Mathematics Tutor Bukit Timah lesson explains how to turn principal trigonometric answers into general solutions and then restrict them to an interval. We will solve sine, cosine and tangent equations with transformed arguments such as 2x or 3x, check signs on the unit circle, avoid discarding roots through division and reject extra roots introduced by squaring. The first answer is short; the 44 original clinics below provide systematic transfer practice.

For JC2 H2 Mathematics learners in Bukit Timah, trigonometric equations are part of the assumed Additional Mathematics background and connect to equation solving in the 2027 H2 Mathematics 9758 syllabus. This guide uses radians, exact multiples of π and the original equation for verification. Its goal is not to replace the wider trigonometry curriculum, but to teach a specific mistake-resistant process before a student works on longer mixed A-Level questions.

The quick answer: general family first, interval filter second

sin θ = sin α: θ=α+2πn or θ=π−α+2πn, where n is any integer.
cos θ = cos α: θ=α+2πn or θ=−α+2πn.
tan θ = tan α: θ=α+πn wherever tangent is defined.
For θ=kx: rewrite the interval for θ. When 0≤x<2π and k is a positive integer, 0≤θ<2kπ.
Recover x: divide every valid θ by k and reject any value outside the original x interval.
Always substitute back: squaring or dividing by an expression that may be zero can change the solution set.

One complete worked question: sin(2x)=1/2

Find all solutions to sin(2x)=1/2 for 0≤x<2π. Write θ=2x. As x ranges from 0 to just below 2π, θ ranges from 0 to just below 4π, covering two complete sine cycles. In the first cycle, sin θ=1/2 at θ=π/6 and θ=5π/6. Add 2π to obtain the second-cycle solutions θ=13π/6 and θ=17π/6.

Divide by two to recover the requested x-values: x=π/12, 5π/12, 13π/12, 17π/12. All four lie in [0,2π), and direct substitution confirms that sin(2x)=1/2. A student who gives only π/12 and 5π/12 has missed a full cycle; a student who gives π/6 and 5π/6 has found θ rather than x.

Why the principal inverse angle is not the whole answer

An inverse-sine or inverse-cosine function is defined with a restricted output range so it can act as a function. The original trigonometric functions are periodic, which means multiple angles share the same sine, cosine or tangent value. A principal inverse-angle output is a starting point. You must construct all valid branches and periods and then filter against the interval in the question.

For positive sine, solutions in a first full cycle occupy quadrants I and II, whereas negative sine occupies III and IV. Positive cosine occupies I and IV, and negative cosine occupies II and III. Tangent repeats every π, so one family with period π is enough. These sign patterns are more reliable than trying to remember a list of disconnected answer pairs.

Sine solutions are a two-branch family

For sin θ=1/2, the reference angle is π/6. One branch is θ=π/6+2πn and the other is θ=5π/6+2πn. For sin θ=−1/2, the first full cycle gives θ=7π/6 and 11π/6. A calculator’s principal arcsin output may be negative, but translating that into the requested non-negative interval is part of the equation-solving task.

A special case is sin θ=0. There are two zeros in the half-open interval [0,2π): θ=0 and θ=π. On the closed interval [0,2π], θ=2π is also included. The equality sign at an interval endpoint is therefore mathematically meaningful, not merely punctuation.

Cosine and the risk of duplicating special roots

The standard cosine family is θ=±α+2πn for a chosen α satisfying cos α equal to the target value. For cos θ=1/2, the solutions in [0,2π) are π/3 and 5π/3. But when cos θ=1, the two signs point to the same periodic family θ=2πn. Listing each twice is not a second solution. For cos θ=−1 the solutions are θ=π+2πn.

When solving cos(3x)=1 over [0,2π), θ=3x covers three full cycles and reaches 0,2π,4π as matching θ values. Thus x=0,2π/3,4π/3. The number of distinct roots is three, not the six that might be guessed from a rule that every cosine value has two solutions per cycle.

Tangent: one branch, shorter period, domain restrictions

For tan θ=1, θ=π/4+πn. In [0,2π), this yields π/4 and 5π/4. For tan(2x)=1 in [0,2π), θ spans two full 2π cycles, giving x=π/8, 5π/8, 9π/8 and 13π/8. The tangent function itself has period π, so there are four distinct solutions in the two-cycle argument range.

Because tan θ=sin θ/cos θ, it is undefined when cos θ=0. Any answer that makes a tangent denominator zero must be excluded from an equation originally written using tan. Conversely, dividing a different equation by cos θ may discard solutions of that original equation if cos θ=0 was allowed. Keep the original expression visible.

A dangerous square: sin x=cos x

The original equation sin x=cos x has solutions x=π/4 and 5π/4 in 0≤x<2π. Squaring both sides gives sin²x=cos²x, which also admits x=3π/4 and 7π/4. Those additional candidates have sine and cosine of equal magnitude but opposite signs, so they fail the original unsquared equation. Squaring is not a reversible step unless the missing sign information is restored.

A useful habit is to label a transformed equation with the words “candidate generator”. After a non-reversible operation such as squaring, use it to generate possible roots, then substitute each candidate into the original. The final answer set must satisfy the starting equation, not just the most recent line of algebra.

Dividing can erase an entire factor

Consider sin x cos x=cos x. Move all terms to one side to obtain cos x(sin x−1)=0. One factor gives cos x=0, so x=π/2 or 3π/2. The other gives sin x=1, producing π/2 again. The complete set is π/2 and 3π/2. If you divide by cos x at the beginning, you lose the valid cos x=0 branch and miss 3π/2.

This mistake is closely related to losing equilibrium solutions when separating a differential equation: dividing by a potentially zero expression excludes exactly the values that may solve the original problem. Factor first, test excluded values, then simplify on the permitted domain.

Use the count as an audit, not as a substitute for solving

On 0≤x<2π, an equation sin(kx)=1/2 has 2k roots for each positive integer k. Cos(kx)=1/2 also has 2k, whereas cos(kx)=1 has only k. Tan(kx)=1 has 2k because tangent repeats every π. These counts help detect missing periods, but a question requiring exact roots still needs the x-values themselves and the given interval.

Changing an interval to a closed endpoint can add a root at x=2π in special cases such as sin(kx)=0, cos(kx)=1 or tan(kx)=0. The half-open interval convention in the clinics below is deliberate: it avoids counting the same angle at both the start and end of one complete x cycle.

Forty-four original exact-solution clinics

All practice questions below ask for all x in 0≤x<2π, with angles in radians and positive integer multipliers. Each gives the argument interval, the first-cycle branches and the final exact solutions. There are no fictional marks or hidden calculator results; the sets are generated from known unit-circle identities and can be checked by substitution.

1. sin(x)=1/2

Try the exact equation. Solve sin(x)=1/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/6 and 5π/6. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/6, 5π/6. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

Transfer and error check. Look at the number of argument cycles. Increasing k compresses more oscillations into the same x interval, which is why extra answers appear. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

2. sin(x)=−1/2

Try the exact equation. Solve sin(x)=−1/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 7π/6 and 11π/6. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=7π/6, 11π/6. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

Transfer and error check. An inverse-trigonometric principal value alone does not describe every matching angle. The second branch or extra periods must be considered. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

3. sin(x)=√3/2

Try the exact equation. Solve sin(x)=√3/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 2π/3. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/3, 2π/3. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

Transfer and error check. The sign of the target determines which unit-circle quadrants are possible. A sign check often catches a wrong reference angle. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

4. sin(x)=−√3/2

Try the exact equation. Solve sin(x)=−√3/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 4π/3 and 5π/3. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=4π/3, 5π/3. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

Transfer and error check. Make sure you have solved for x, not just the temporary argument θ=kx. Forgetting to divide by k leaves roots on the wrong scale. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

5. sin(x)=0

Try the exact equation. Solve sin(x)=0 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 0 and π. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=0, π. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

Transfer and error check. Every final answer must lie below 2π because the upper endpoint is excluded. A value equal to 2π would be rejected in this exercise. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

6. cos(x)=1/2

Try the exact equation. Solve cos(x)=1/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 5π/3. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/3, 5π/3. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

Transfer and error check. If the target is an extreme cosine value, matching branches may collapse into one distinct family rather than two. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

7. cos(x)=−1/2

Try the exact equation. Solve cos(x)=−1/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 2π/3 and 4π/3. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=2π/3, 4π/3. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

Transfer and error check. For tangent, respect the π period. Applying a second, mirrored sine branch would create duplication or false solutions. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

8. cos(x)=√2/2

Try the exact equation. Solve cos(x)=√2/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/4 and 7π/4. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/4, 7π/4. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

Transfer and error check. Use direct substitution of one non-obvious root to test that the correct function and sign were retained throughout. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

9. cos(x)=−√2/2

Try the exact equation. Solve cos(x)=−√2/2 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 3π/4 and 5π/4. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=3π/4, 5π/4. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

Transfer and error check. Sort solutions from smallest to largest. This makes missing cycles and duplicate roots easier to see. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

10. cos(x)=0

Try the exact equation. Solve cos(x)=0 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/2 and 3π/2. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/2, 3π/2. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

Transfer and error check. A root count is a useful audit, but the final x-values carry the precise mathematical information the question requests. If the interval were changed to 0≤x≤2π, test the new endpoint independently instead of automatically adding it. Finally explain what would happen if k increased by one while the original half-open x interval stayed fixed.

11. cos(x)=1

Try the exact equation. Solve cos(x)=1 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 0. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=0. There is 1 distinct solution in [0,2π). All belong to the original interval and must satisfy the original cos equation.

12. cos(x)=−1

Try the exact equation. Solve cos(x)=−1 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π. There is 1 distinct solution in [0,2π). All belong to the original interval and must satisfy the original cos equation.

13. tan(x)=1

Try the exact equation. Solve tan(x)=1 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/4 and 5π/4. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/4, 5π/4. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

14. tan(x)=−1

Try the exact equation. Solve tan(x)=−1 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 3π/4 and 7π/4. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=3π/4, 7π/4. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

15. tan(x)=√3

Try the exact equation. Solve tan(x)=√3 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 4π/3. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=π/3, 4π/3. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

16. tan(x)=−√3

Try the exact equation. Solve tan(x)=−√3 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 2π/3 and 5π/3. Repeating those angles through 1 full θ cycle and dividing by 1 gives x=2π/3, 5π/3. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

17. tan(x)=0

Try the exact equation. Solve tan(x)=0 for 0≤x<2π. First set θ=x; the argument then runs through 0≤θ<2π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

18. sin(2x)=1/2

Try the exact equation. Solve sin(2x)=1/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/6 and 5π/6. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/12, 5π/12, 13π/12, 17π/12. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

19. sin(2x)=−1/2

Try the exact equation. Solve sin(2x)=−1/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 7π/6 and 11π/6. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=7π/12, 11π/12, 19π/12, 23π/12. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

20. sin(2x)=√3/2

Try the exact equation. Solve sin(2x)=√3/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 2π/3. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/6, π/3, 7π/6, 4π/3. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

21. sin(2x)=−√3/2

Try the exact equation. Solve sin(2x)=−√3/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 4π/3 and 5π/3. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=2π/3, 5π/6, 5π/3, 11π/6. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

22. sin(2x)=0

Try the exact equation. Solve sin(2x)=0 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 0 and π. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=0, π/2, π, 3π/2. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

23. cos(2x)=1/2

Try the exact equation. Solve cos(2x)=1/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 5π/3. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/6, 5π/6, 7π/6, 11π/6. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

24. cos(2x)=−1/2

Try the exact equation. Solve cos(2x)=−1/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 2π/3 and 4π/3. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/3, 2π/3, 4π/3, 5π/3. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

25. cos(2x)=√2/2

Try the exact equation. Solve cos(2x)=√2/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/4 and 7π/4. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/8, 7π/8, 9π/8, 15π/8. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

26. cos(2x)=−√2/2

Try the exact equation. Solve cos(2x)=−√2/2 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 3π/4 and 5π/4. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=3π/8, 5π/8, 11π/8, 13π/8. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

27. cos(2x)=0

Try the exact equation. Solve cos(2x)=0 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/2 and 3π/2. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/4, 3π/4, 5π/4, 7π/4. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

28. cos(2x)=1

Try the exact equation. Solve cos(2x)=1 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 0. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=0, π. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

29. cos(2x)=−1

Try the exact equation. Solve cos(2x)=−1 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/2, 3π/2. There are 2 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

30. tan(2x)=1

Try the exact equation. Solve tan(2x)=1 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/4 and 5π/4. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/8, 5π/8, 9π/8, 13π/8. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

31. tan(2x)=−1

Try the exact equation. Solve tan(2x)=−1 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 3π/4 and 7π/4. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=3π/8, 7π/8, 11π/8, 15π/8. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

32. tan(2x)=√3

Try the exact equation. Solve tan(2x)=√3 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 4π/3. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/6, 2π/3, 7π/6, 5π/3. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

33. tan(2x)=−√3

Try the exact equation. Solve tan(2x)=−√3 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 2π/3 and 5π/3. Repeating those angles through 2 full θ cycles and dividing by 2 gives x=π/3, 5π/6, 4π/3, 11π/6. There are 4 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original tan equation.

34. tan(2x)=0

Try the exact equation. Solve tan(2x)=0 for 0≤x<2π. First set θ=2x; the argument then runs through 0≤θ<4π. Decide which unit-circle angles have the stated tan value, and keep each branch through every period covered by the new argument range.

35. sin(3x)=1/2

Try the exact equation. Solve sin(3x)=1/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/6 and 5π/6. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

36. sin(3x)=−1/2

Try the exact equation. Solve sin(3x)=−1/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 7π/6 and 11π/6. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=7π/18, 11π/18, 19π/18, 23π/18, 31π/18, 35π/18. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

37. sin(3x)=√3/2

Try the exact equation. Solve sin(3x)=√3/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 2π/3. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=π/9, 2π/9, 7π/9, 8π/9, 13π/9, 14π/9. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

38. sin(3x)=−√3/2

Try the exact equation. Solve sin(3x)=−√3/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 4π/3 and 5π/3. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=4π/9, 5π/9, 10π/9, 11π/9, 16π/9, 17π/9. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

39. sin(3x)=0

Try the exact equation. Solve sin(3x)=0 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated sin value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 0 and π. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=0, π/3, 2π/3, π, 4π/3, 5π/3. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original sin equation.

40. cos(3x)=1/2

Try the exact equation. Solve cos(3x)=1/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/3 and 5π/3. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=π/9, 5π/9, 7π/9, 11π/9, 13π/9, 17π/9. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

41. cos(3x)=−1/2

Try the exact equation. Solve cos(3x)=−1/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 2π/3 and 4π/3. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=2π/9, 4π/9, 8π/9, 10π/9, 14π/9, 16π/9. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

42. cos(3x)=√2/2

Try the exact equation. Solve cos(3x)=√2/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/4 and 7π/4. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=π/12, 7π/12, 3π/4, 5π/4, 17π/12, 23π/12. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

43. cos(3x)=−√2/2

Try the exact equation. Solve cos(3x)=−√2/2 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are 3π/4 and 5π/4. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=π/4, 5π/12, 11π/12, 13π/12, 19π/12, 7π/4. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

44. cos(3x)=0

Try the exact equation. Solve cos(3x)=0 for 0≤x<2π. First set θ=3x; the argument then runs through 0≤θ<6π. Decide which unit-circle angles have the stated cos value, and keep each branch through every period covered by the new argument range.

Correct complete solution. In the first θ cycle the matching angles are π/2 and 3π/2. Repeating those angles through 3 full θ cycles and dividing by 3 gives x=π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6. There are 6 distinct solutions in [0,2π). All belong to the original interval and must satisfy the original cos equation.

Ten shorter challenges without the topic signposted

1. sin x=1/2 on [0,2π): π/6, 5π/6.
2. cos x=1/2: π/3, 5π/3.
3. tan x=1: π/4, 5π/4.
4. sin 2x=0: 0, π/2, π, 3π/2.
5. cos 2x=1: 0, π.
6. tan 2x=0: 0, π/2, π, 3π/2.
7. sin x=cos x: π/4, 5π/4.
8. sin x cos x=cos x: π/2, 3π/2.
9. On the closed interval [0,2π], sin x=0 includes 0, π and 2π.
10. cos 3x=1 on [0,2π): 0, 2π/3, 4π/3.

The one-minute diagnostic and three-person tutorial

Ask one student why sin(2x)=1/2 gives four roots instead of two; ask another student why cos(2x)=1 gives only two roots; ask a third to check all values against 0≤x<2π. Then switch the functions. A learner who can explain the two-solution versus one-solution distinction through the unit circle has a much stronger foundation than one who memorises a button sequence.

A useful diagnosis is to change exactly one thing at a time. If the learner confuses π/6 with π/12, change the multiplier k. If they leave out 13π/12 and 17π/12, change the argument interval. If they include 2π, change the interval endpoint symbol. If they accept 3π/4 for sin x=cos x, ask them to substitute it into the original. Each error points to a specific mathematical repair.

A five-day revision path

Day one: draw the sine, cosine and tangent reference-angle branches on a unit circle. Day two: construct general solutions with the correct periods and integer n. Day three: introduce kx and rewrite the argument interval before solving. Day four: practise equations where division by sine or cosine might erase roots. Day five: solve mixed problems and check exact endpoints, including one with an inclusive upper bound. Each day should end with a one-sentence explanation of why the final set is complete.

Parents can ask the learner to describe how many cycles the argument traverses before reviewing the actual calculations. Tutors can ask why a particular candidate has been excluded. Neither requires a separate paid calculator tool. In a three-pupil learning format, one student can own the unit-circle branch, another can own the transformed interval, and a third can audit the final set, then rotate roles.

Frequently asked questions

Why does sin(2x)=1/2 have four solutions?

The argument 2x spans two sine cycles while x runs from 0 to 2π. Each cycle contributes two distinct solutions for the target 1/2.

Does cosine always have two solutions per period?

Not for extreme target values such as 1 and −1, which occur once per 2π period in a half-open interval. General counts need this exception.

Why does tangent have period π?

Because the sine and cosine both change sign when an angle increases by π, leaving their ratio unchanged wherever cosine is nonzero.

Is a calculator principal value wrong?

No. It is one defined inverse-function output. It is incomplete if the question requests all periodic solutions.

Can squaring create false solutions?

Yes. It can erase sign information. Substitute every candidate back into the original unsquared equation.

Can division lose solutions?

Yes. Dividing by an expression that may equal zero can remove a branch where that factor is zero. Factor and check zero cases before dividing.

Should H2 trigonometry be in radians?

Use whatever angular unit the problem states. This guide works in radians and uses exact multiples of π, consistent with advanced mathematics practice.

How does this relate to the H2 9758 syllabus?

The 2027 H2 syllabus assumes trigonometric functions, identities and equations from Additional Mathematics and includes equation-solving techniques and modelling. This article isolates the general-solution-to-interval skill rather than asserting that every example is a verbatim past examination question.

Continue learning and check the official syllabus

Normal Conditional Tail Probabilities
Trigonometric General Solutions and Interval Checks: current guide
Maclaurin Quotient Series and Missing Coefficients
Infinite Geometric Series and Convergence

For fuller topic coverage, see the JC H2 trigonometric identities and equations guide and the Bukit Timah JC2 tuition hub. The immutable Secondary 1 Mathematics Tutor Clementi small-group reference illustrates a teaching format at another level and locality.

Official curriculum reference: SEAB 2027 H2 Mathematics 9758 syllabus, Assumed Knowledge: Trigonometric Functions, Identities and Equations, and H2 equation-solving. All exercises above are original mathematical constructs, not official examination questions or actual student outcomes.

JC2 H2 Mathematics Tutor Bukit Timah | Trigonometric General Solutions and Interval Checks