Trigonometric equations containing different multiples of one angle can be solved systematically without guessing from a calculator. Rewrite cosine as a sine using a complementary angle, apply the two general families for sin A = sin B, and then filter every candidate through the requested interval. The challenge is not usually the algebra: it is counting all valid periodic solutions without overlooking one or reporting duplicates.
For example, solve sin 3θ = cos 2θ for 0° ≤ θ < 360°. Because cos 2θ = sin(90°−2θ), the equation has two families. One gives 5θ=90°+360°n, or θ=18°+72°n. The other gives θ=90°+360°n. On the stated interval, the complete distinct answers are **18°, 90°, 162°, 234° and 306°**. The 90° solution occurs in both families but must be counted only once.
Did you know that this duplicated angle is also a tangential meeting point of the two trigonometric graphs? This guide explains the identities, validates every solution, shows an equivalent product-to-sum method and develops a reliable approach for different angle multipliers and strict interval endpoints.
Read the argument of each trigonometric function
The first term is sin(3θ), the sine of three times θ. The second is cos(2θ), the cosine of twice θ.
These are not sin³θ and cos²θ. Powers of the function value and multiples of the angle are completely different operations.
The problem specifies one real angle θ, but the two expressions oscillate at different frequencies as θ varies. Each can equal the other at several different positions in one complete θ-revolution.
A careful solution keeps the bracketed arguments 3θ and 2θ intact and uses periodic general solutions rather than a single inverse-trigonometric principal value. This is how all relevant angles are recovered.
Identify the angle interval before solving
The requested domain is 0° ≤ θ < 360°. It includes zero but excludes 360°.
After deriving a general solution, we must list every integer n that produces an angle inside that interval and reject values outside it.
The interval is half-open to avoid double-counting the same direction at both ends of a revolution. However, not every equation has 0° or 360° as a solution.
The interval symbols must still be read exactly. A student who writes a memorised list of special angles without applying the stated bounds may include invalid endpoints or omit legitimate values near the upper edge.
Use the complementary-angle identity
The cofunction identity is cos α = sin(90°−α) in degrees, for any real α.
Apply it to the right-hand side with α=2θ. Then cos2θ=sin(90°−2θ).
Thus the original equation becomes sin3θ=sin(90°−2θ).
We have changed the equation into equality of two sine values, allowing one standard general-solution rule to handle both sides.
It is not necessary to make 90°−2θ positive. Trigonometric functions are defined for negative angles, and the general solution formulas already account for coterminal directions.
Recall the two sine-equality families
If sin A = sin B, then the angles A and B need not be identical.
The complete general solutions in degrees are A=B+360°n or A=180°−B+360°n, where n is any integer.
The first family captures angles with the same directed sine phase. The second captures their reflection around the 90° direction, which preserves sine.
Both families are required. Using only A=B+360°n can omit valid values, while inventing a plus-or-minus rule A=±B+360°n would generally be incorrect for sine.
Here we set A=3θ and B=90°−2θ and work through both families separately.
Solve the first family algebraically
Use 3θ=(90°−2θ)+360°n. Add 2θ to both sides.
The result is 5θ=90°+360°n. Divide by five to obtain θ=18°+72°n.
This is one infinite family of exact real solutions indexed by integer n.
The 72° step comes from 360°/5. It does not mean the original equation itself has a period of 72°; it means this particular family produces a sequence of solutions separated by 72°.
The actual equation and its graph can have additional solutions from the second family, and those need separate consideration.
List the first family’s permitted angles
Apply 0° ≤ 18°+72°n < 360°.
At n=0 the angle is 18°. At n=1 it is 90°. At n=2 it is 162°. At n=3 it is 234°. At n=4 it is 306°.
The value n=5 would give 378°, outside the upper bound. The value n=−1 would give −54°, below the lower bound.
Thus the first family contributes exactly five permitted angles: 18°, 90°, 162°, 234° and 306°. The next step is to determine what the second family contributes beyond this list.
Solve the second family correctly
Use 3θ=180°−(90°−2θ)+360°n.
Carefully simplify the bracket: 180°−90°+2θ=90°+2θ. Therefore 3θ=90°+2θ+360°n.
Subtract 2θ from both sides to obtain θ=90°+360°n.
This is the second general family. On the interval 0° ≤ θ < 360°, only n=0 is allowed, yielding θ=90°.
The calculation shows why the second family cannot be ignored merely because the first one already contains several answers. In this particular example its permitted member happens to overlap with one of them.
Remove duplicate angles from the union
The first family yielded 18°, 90°, 162°, 234° and 306°.
The second family yielded 90°. That angle was already present, so it must not be counted twice.
The final set of distinct solutions is {18°,90°,162°,234°,306°}. There are five values, not six.
A solution set counts distinct permissible angles. Two different algebraic derivations of the same angle do not create two mathematical answers. This distinction becomes particularly important at a tangency or at a repeated root of a transformed trigonometric equation.
Check the eighteen-degree solution exactly
At θ=18°, the left side is sin(3×18°)=sin54°.
The right side is cos(2×18°)=cos36°. The complementary-angle identity gives sin54°=cos36°, because 54°+36°=90°.
Therefore 18° is an exact solution. No decimal evaluation is necessary.
The same complementary-angle logic explains the first family generally. Its members make the sine argument 3θ coterminal with the complementary angle 90°−2θ, so the two outputs agree exactly.
Check the ninety-degree solution
At θ=90°, the left side is sin270°=−1.
The right side is cos180°=−1. Both are exactly equal.
This angle belongs to the first family with n=1 because 18°+72°=90°. It also belongs to the second family with n=0.
The duplicate is therefore real and mathematically meaningful, not a sign that one algebraic branch must be discarded. We keep the angle once in the solution set and recognise that both sine-equality mechanisms coincide there.
Check the one-hundred-sixty-two-degree solution
At θ=162°, the left side is sin486°.
Subtract one full turn: sin486°=sin126°. By supplementary-angle symmetry, sin126°=sin54°.
The right side is cos324°=cos36°, since cosine is symmetric around 360°. And sin54°=cos36°.
Therefore both sides agree exactly. The verification uses only standard angle identities, not a calculator rounded to a few decimal places. It confirms that the solution near the middle of the interval is legitimate.
Check the two-hundred-thirty-four-degree solution
At θ=234°, the left side is sin702°=sin342° after subtracting 360°.
Since sin342°=−sin18°, the left side is negative. The right side is cos468°=cos108°.
Use cos108°=−cos72° and sin18°=cos72°. Therefore the right side is −sin18°, matching the left side.
Thus 234° is valid. This check is helpful because both sides are negative; students who use only acute-angle intuition can mistakenly discard third- or fourth-quadrant solutions.
Check the three-hundred-six-degree solution
At θ=306°, the left side is sin918°.
Subtract 720° to obtain sin198°=−sin18°. The right side is cos612°=cos252°.
Since cos252°=−cos72° and cos72°=sin18°, the right side is also −sin18°.
Therefore 306° satisfies the equation. All five reported angles now have exact trigonometric support, and the integer-family filtering has established that none was omitted from the requested interval.
Why a principal inverse value is not sufficient
If a student uses a calculator to solve sin3θ=cos2θ numerically, the device may return one root close to 18°.
That does not prove there is only one solution. The sine and cosine arguments move through several cycles while θ moves from 0° to 360°.
A principal inverse-sine value is only one representative of an entire periodic family. The reflected sine family can provide another sequence as well.
The general-solution approach records all possibilities symbolically, then selects the permitted ones. A calculator can check values but should not replace the interval-completeness reasoning.
Use product-to-sum as an independent method
Rewrite the equation as sin3θ−cos2θ=0. Replace cosine by sine: sin3θ−sin(90°−2θ)=0.
Use the identity sin A−sin B=2cos[(A+B)/2]sin[(A−B)/2].
With A=3θ and B=90°−2θ, the equation becomes 2cos[(θ+90°)/2]sin[(5θ−90°)/2]=0.
A product is zero when either factor is zero. Solving the sine factor gives θ=18°+72°n. Solving the cosine factor gives θ=90°+360°n.
The two solution families match the complementary-angle method exactly. This provides an elegant cross-check without evaluating the original functions numerically.
Solve the first product factor
From sin[(5θ−90°)/2]=0, require (5θ−90°)/2=180°n for integer n.
Multiply by two: 5θ−90°=360°n. Therefore θ=18°+72°n.
This is the same five-angle family found earlier.
The factorisation method can be useful when a problem asks why two sets of angular solutions arise. It transforms equality of two trigonometric functions into a product with two distinct zero conditions, each corresponding to a mathematically valid branch.
Solve the cosine product factor
From cos[(θ+90°)/2]=0, require (θ+90°)/2=90°+180°n.
Multiply by two: θ+90°=180°+360°n, giving θ=90°+360°n.
On the standard half-open revolution interval, this contributes θ=90°.
The overlap with the first factor occurs because both factors vanish at 90°. A product of two zero factors is still one zero at one input, not two distinct angle values.
The algebra therefore explains the duplicate in an especially transparent way.
The duplicate angle is a tangential graph meeting
Define F(θ)=sin3θ−cos2θ, now treating θ as a radian variable for calculus. The corresponding duplicated solution is θ=π/2.
Differentiate: F′(θ)=3cos3θ+2sin2θ. At θ=π/2, both terms vanish, so F′(π/2)=0.
The second derivative is F″(θ)=−9sin3θ+4cos2θ. At π/2, it equals 9−4=5>0.
Thus F has a local minimum of zero at this angle rather than changing sign through it. The two original graphs meet with the same local gradient there, which explains why the algebraic solution families coincide. This is calculus enrichment, not a requirement for listing the angles.
A repeated root does not give another distinct solution
The product-to-sum expression has two factors that both vanish at θ=90°. That is analogous to a function having a repeated zero at the same input.
In a list of real solutions, an input appears once regardless of how many factors vanish there.
This is different from counting multiplicities in polynomial factorisation, where a repeated root may have algebraic significance. A solution set records distinct angles.
Students who report “90° twice” because each family produces it have not misunderstood the trigonometric identity, but have confused multiplicity with membership of a set. Deduplicating is an essential final step.
The original equation’s shared period
The function sin3θ has period 120°, since increasing θ by 120° increases its argument by 360°.
The function cos2θ has period 180°. A common period of both is 360°, because three cycles of the first and two cycles of the second fit inside that interval.
Therefore the difference sin3θ−cos2θ repeats after 360°. The five distinct solutions in one half-open revolution reproduce at shifts of 360°.
The first family happens to be spaced by 72°, but that does not make 72° a period of the entire equation. A solution spacing and a function period are different concepts.
What happens over two revolutions?
If the domain is 0° ≤ θ < 720°, the equation completes two common periods.
The five distinct solutions in the first revolution are 18°,90°,162°,234°,306°. Add 360° to each for the second revolution: 378°,450°,522°,594°,666°.
Thus there are ten distinct solutions in the two-period interval.
The duplicated second-family angle also recurs at 450°, but it is already included in the shifted first family. Count the union of distinct values rather than adding a fixed count from each general family blindly.
A restricted first-half interval
Suppose the question instead asks for 0° ≤ θ < 180°.
From the complete list, only 18°,90° and 162° lie in this interval.
The values 234° and 306° are valid general solutions but outside the new permitted domain, so they must be excluded.
This illustrates the final filtering step. The algebraic general families stay unchanged; only the selection of integer n-values changes with the angle interval. A correct response preserves every constraint from the question without altering the trigonometric equation itself.
A strict interval below ninety degrees
If the permitted interval is 0° < θ < 90°, the only solution from the full list is 18°.
The 90° solution is excluded by the strict upper inequality, and zero is not a solution of this equation anyway.
The remaining values lie above 90° and therefore fail the bound.
This case shows why endpoint symbols matter. Even though 90° has a special role as a duplicated solution, it must be discarded when the question explicitly excludes it. Algebraic importance does not override the permitted domain.
Translate the whole method into radians
In radians, use cos2θ=sin(π/2−2θ). The first sine-equality family is 3θ=π/2−2θ+2πn.
Thus 5θ=π/2+2πn, giving θ=π/10+(2π/5)n.
The second family is 3θ=π−(π/2−2θ)+2πn=π/2+2θ+2πn, giving θ=π/2+2πn.
On 0≤θ<2π, the five distinct answers are π/10, π/2, 9π/10, 13π/10 and 17π/10. The angle π/2 appears in both families and is included only once.
Convert exact radian angles back to degrees
Multiply a radian value by 180°/π to convert to degrees.
For example, π/10 corresponds to 18°. The value 9π/10 corresponds to 162°, and 13π/10 corresponds to 234°.
The radian solution list π/10, π/2, 9π/10, 13π/10 and 17π/10 is therefore exactly equivalent to the degree list 18°,90°,162°,234°,306°.
A mixed solution that uses 360 in a radian periodic formula or 2π in a degree equation is inconsistent. Choose one angular unit at the start and preserve it throughout the general-solution and interval-filtering steps.
Another example: solve sin 2θ = sin θ
Consider sin2θ=sinθ on 0°≤θ<360°. The general sine-equality families give 2θ=θ+360°n or 2θ=180°−θ+360°n.
The first yields θ=360°n, so θ=0° in the interval. The second yields 3θ=180°+360°n, hence θ=60°+120°n.
The permitted second-family values are 60°,180° and 300°.
Thus the complete solution set is {0°,60°,180°,300°}. It has four distinct angles. The same two-family logic works even when neither side is a cosine.
A cosine-equality equation follows a different rule
Suppose the task is cos3θ=cosθ rather than an equality of two sine values.
For cosine, the complete general families are A=B+360°n or A=−B+360°n.
Apply them: 3θ=θ+360°n gives θ=180°n, while 3θ=−θ+360°n gives θ=90°n.
On 0°≤θ<360°, the union is {0°,90°,180°,270°}. The first family is already contained in the second.
Do not apply the sine reflection formula A=180°−B to a cosine equality without conversion. Different trigonometric functions have different symmetry identities.
Another mixed example: sin 3θ = cos θ
Rewrite cosθ as sin(90°−θ), then solve sin3θ=sin(90°−θ).
The first family gives 3θ=90°−θ+360°n, so θ=22.5°+90°n. Its permitted values are 22.5°,112.5°,202.5°,292.5°.
The second family gives 3θ=90°+θ+360°n, so θ=45°+180°n. Its permitted values are 45° and 225°.
There is no overlap between these two lists on the standard interval. Therefore this equation has six distinct solutions, illustrating that the duplicate in sin3θ=cos2θ was a special property of that pair of multipliers.
Generalise to sin mθ = cos nθ
Let m and n be positive integers, and work in radians. Use cos(nθ)=sin(π/2−nθ).
The first sine-equality family gives mθ=π/2−nθ+2πk, so θ=[π/2+2πk]/(m+n).
The second gives mθ=π/2+nθ+2πk, so (m−n)θ=π/2+2πk.
When m≠n, the second family is θ=[π/2+2πk]/(m−n). When m=n, that second equation becomes 0=π/2+2πk, which has no integer k-solution. Every derived angle must then be filtered into the problem’s permitted interval and duplicates removed.
Why the difference m minus n can be negative
If m The integer k can be negative or positive, and the resulting θ may still lie inside the permitted interval after calculation. For example, reversing the multipliers from sin3θ=cos2θ to sin2θ=cos3θ changes the second family, and additional solutions may appear. A student should not discard a family solely because a denominator or intermediate angle is negative. Solve its integer cases and then apply the actual θ-domain rather than imposing an unstated positivity condition on intermediate expressions. The sine graph is symmetric about a vertical line through each maximum. Within one revolution, a typical sine level is reached once on each side of a peak. The identity sinA=sinB therefore permits both A congruent to B and A congruent to 180°−B modulo a full turn. At the extreme sine levels ±1, these descriptions can coincide. That is one reason duplicate angles may appear after solving both families. Understanding the graph symmetry helps students remember the formula rather than treating it as arbitrary notation. It also supplies a qualitative solution-count check before any long list of integers is prepared. Replacing cos2θ with sin(90°−2θ) is an identity valid for every real angle. It neither loses nor creates solutions. Applying the complete sine-equality rule is also reversible: every resulting family genuinely satisfies the sine equality. The only later operation is restricting θ to the stated interval and deduplicating repeated values. Neither involves a risky division by a trigonometric expression that might be zero. This is a strength of the method. Alternative approaches that divide by sinθ or cosθ may accidentally discard angles where the denominator vanishes. A transformation based on identities and general solutions avoids that particular hazard. At θ=18°, the equality reduces to sin54°=cos36°, an exact complementary relation. At θ=90°, both sides equal −1. At the other listed angles, ordinary periodic and supplementary-angle identities verify the same equality. A calculator may show matching decimals to several places, but numerical agreement cannot prove that there are no additional roots. Exact identities establish validity of the candidates, while general-solution families and interval filtering establish completeness. Both parts are needed for a fully justified trigonometric equation solution. Define F(θ)=sin3θ−cos2θ. The original equation asks where F equals zero. Over a complete 360° revolution, the graph of F returns to its starting shape and has five distinct zeros. At 90°, the graph touches the horizontal axis rather than crossing it, while the other roots are crossings. This explains why a numerical graph may produce an unexpected number of visible sign changes: a tangential zero does not change the sign of F. The general algebra detects both crossing and touching roots. A sign-change-only numerical search could miss the tangential 90° solution unless it also checks stationary zero contacts. First record the angle interval and angular unit. Second convert cosine to sine using cosα=sin(90°−α) in degrees or cosα=sin(π/2−α) in radians. Third write both general sine-equality families. Fourth solve each resulting linear equation in θ, retaining the integer parameter. Fifth list every candidate in the permitted interval, including strict endpoint checks. Sixth take the union of distinct values and verify representative answers in the original trigonometric equation. For sin3θ=cos2θ on 0°≤θ<360°, this procedure gives 18°,90°,162°,234°,306° with no missing or duplicated values. Question one: solve sin3θ=cos2θ on 0°≤θ<360°. Answer: 18°,90°,162°,234°,306°. Question two: solve the same equation on 0°≤θ<180°. Answer: 18°,90°,162°. Question three: solve it on 0°<θ<90°. Answer: 18° only. Question four: solve sin2θ=sinθ on 0°≤θ<360°. Answer: 0°,60°,180°,300°. Question five: solve cos3θ=cosθ on the same interval. Answer: 0°,90°,180°,270°. Question six: solve sin3θ=cosθ on the same interval. Answer: 22.5°,45°,112.5°,202.5°,225°,292.5°. Question seven: give the radian solutions to Question one on 0≤θ<2π. Answer: π/10,π/2,9π/10,13π/10,17π/10. For each, show both angle-equality families and remove duplicates only after checking the interval. Solving equations with different trigonometric angle multipliers combines cofunction identities, sine and cosine symmetries, periodicity, angle intervals and exact checking. The tangential-zero calculus interpretation is enrichment beyond the basic angle-listing method. Singapore families can consult SEAB’s applicable 2026 O-Level Additional Mathematics and 2027 SEC G3 syllabus resources according to the learner’s school course. Related eduKateSG guides include How to List Every Trigonometric Solution on an Interval Crossing Zero, How to Solve a Hidden Quadratic Trigonometric Equation, How to Solve Trigonometric Equations Using sin θ + cos θ as One Variable and the Additional Mathematics archive. Families can enquire about suitable eduKate Singapore small-group Mathematics tutorials near Sixth Avenue MRT. The enduring skill is expressing every periodic branch exactly, respecting the specified angle interval and reporting each distinct solution once. Properly taught kids shine a bright light into the future.Why the exact equality sin A = sin B needs both branches
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