A trigonometric equation can be quadratic in disguise when the same sine or cosine expression appears squared and to the first power. Substitute a single letter for that trigonometric quantity, solve the ordinary quadratic, and then return to the angle. The crucial extra check is that sine and cosine must lie between −1 and 1.
For example, 2sin²θ − sinθ − 1 = 0 becomes 2u² − u − 1 = 0 when u = sinθ. Its roots are u = 1 and u = −1/2. On 0° ≤ θ < 360°, the complete solutions are θ = 90°, 210° and 330°. This guide explains every stage, including impossible substituted roots, general-angle solutions, interval endpoints and quadratic inequalities in sine or cosine.
Recognise the repeated trigonometric quantity
The equation 2sin²θ − sinθ − 1 = 0 contains only the quantities (sinθ)², sinθ and a constant. It has the same algebraic structure as 2u² − u − 1 = 0.
The substitution u = sinθ reduces a trigonometric expression to a familiar quadratic. The point is not to change the mathematical problem, but to isolate a pattern we already know how to solve.
A student should distinguish sin²θ, meaning (sinθ)², from sin(2θ), which is the sine of twice the angle. The double-angle expression is not automatically the square of sine.
Once the structure is recognised, solve in u before attempting to list angles. Mixing quadratic factorisation and quadrant reasoning in one step often creates avoidable omissions.
State the range of sine before solving
For every real angle θ, −1 ≤ sinθ ≤ 1. Therefore u = sinθ must satisfy −1 ≤ u ≤ 1.
That restriction is essential. The quadratic 2u² − u − 1 = 0 might produce numbers that do not lie in the sine range in a different example. Such numbers cannot correspond to real angles.
Unlike u = 2^x, where u must be positive, or u = x², where u must be non-negative, a sine substitution permits negative u-values.
Write the inherited range directly under the substitution. It provides an admissibility test before the trigonometric angle stage and avoids treating every algebraic u-root as automatically meaningful.
Factorise the opening quadratic
Put u = sinθ into 2sin²θ − sinθ − 1 = 0. The transformed equation is 2u² − u − 1 = 0.
Factorise to (2u + 1)(u − 1) = 0. Therefore u = −1/2 or u = 1.
Both values lie in [−1,1], so both are admissible sine values. Return to the original meaning: sinθ = −1/2 or sinθ = 1.
These are two separate trigonometric equations. Each must be solved over the complete angle interval requested, and each may contribute a different number of angles.
Find all angles for sine equal to one
On 0° ≤ θ < 360°, sinθ = 1 holds at θ = 90° only.
The unit-circle point is the top of the circle, where the vertical coordinate is 1. No other angle in one half-open complete revolution has the same maximum value.
Therefore the root u = 1 contributes one distinct angle. This is a boundary case of the sine range and should not be forced into the ordinary two-angle pattern.
A student who always expects two angles per valid sine value will overcount maximum or minimum values. The range endpoint u = 1 is special because it occurs only once per full period.
Find all angles for sine equal to negative one-half
The reference angle for |sinθ| = 1/2 is 30°. Sine is negative in Quadrants III and IV.
Therefore sinθ = −1/2 on 0° ≤ θ < 360° at θ = 180° + 30° = 210° and θ = 360° − 30° = 330°.
Both satisfy the equation. They are distinct from the 90° solution belonging to sinθ = 1.
The complete ordered answer is θ = 90°, 210° or 330°. Students should not report only 210°, since a calculator may return a principal negative angle that represents only one of the relevant periodic solutions.
Check all three solutions in the original equation
At θ = 90°, sine equals 1. The left side becomes 2(1)² − 1 − 1 = 0.
At θ = 210° and 330°, sine equals −1/2. The left side becomes 2(1/4) − (−1/2) − 1 = 1/2 + 1/2 − 1 = 0.
Every reported angle makes the original expression zero. A complete verification also asks whether any valid angles were missed, which the two sine cases and full-period range answer.
Checking only the transformed quadratic confirms the allowed sine values but not necessarily the completeness of the angles in the requested interval.
An impossible u-root should be rejected
Consider sin²θ − 3sinθ + 2 = 0. Set u = sinθ to get (u − 1)(u − 2) = 0.
The algebraic roots are u = 1 and u = 2. But sinθ cannot equal 2 for any real angle, so u = 2 must be rejected.
The admissible condition sinθ = 1 yields θ = 90° in 0° ≤ θ < 360°. There is only one solution.
A quadratic with two real roots is not necessarily a trigonometric equation with two sine values. Each root of the substituted quadratic must pass the range check before being converted into angles.
Why a negative sine root is not invalid
For sinθ = −1/2, a negative sine value is fully valid because sine represents a signed vertical unit-circle coordinate.
The valid angles in the full degree interval are 210° and 330°. They lie below the horizontal axis on the unit circle.
Students sometimes confuse the positive side lengths of a reference triangle with the signed value of sine. The reference triangle supplies magnitude; the quadrant supplies sign.
A range condition −1 ≤ u ≤ 1 does not mean u ≥ 0. The wrong positivity assumption would incorrectly erase two of the three solutions in the opening example.
A hidden quadratic in cosine works the same way
Solve 2cos²θ + cosθ − 1 = 0 on 0° ≤ θ < 360°. Set v = cosθ, with −1 ≤ v ≤ 1.
Then 2v² + v − 1 = (2v − 1)(v + 1) = 0. Thus v = 1/2 or v = −1.
For cosθ = 1/2, θ = 60° or 300°. For cosθ = −1, θ = 180°.
Therefore the complete answer is θ = 60°, 180° or 300°. This mirrors the sine method, but uses cosine’s quadrant signs and maximum/minimum locations instead of sine’s.
Do not mix sin²θ with cos²θ without an identity
If an equation contains sin²θ and cos²θ, use sin²θ + cos²θ = 1 when it helps rewrite everything in one trigonometric quantity.
For example, 2sin²θ + cos²θ − 3sinθ = 0 becomes 2sin²θ + (1 − sin²θ) − 3sinθ = 0.
Simplifying gives sin²θ − 3sinθ + 1 = 0, a quadratic in sinθ. Its roots are (3 ± √5)/2.
Only (3 − √5)/2 lies between −1 and 1. The other root is greater than 1 and cannot be a sine value. The valid sine value then determines the angles using quadrant reasoning.
An equation may require the quadratic formula
Consider 2sin²θ − sinθ − 1/4 = 0. Put u = sinθ and multiply by four: 8u² − 4u − 1 = 0.
The quadratic formula gives u = [4 ± √(16 + 32)]/16 = [4 ± 4√3]/16 = (1 ± √3)/4.
Both values lie in [−1,1], so both are admissible. The positive value produces two sine angles in Quadrants I and II; the negative value produces two in Quadrants III and IV.
Exact inverse-sine expressions may be suitable when the question asks for exact forms, while degree approximations may be used to the requested accuracy. Do not round u before deciding which quadrants apply.
General solutions prevent missed repetitions
For sinθ = s with −1 < s < 1, a reference-angle approach gives two families of solutions per full period.
If α is a principal angle satisfying sinα = s, one convenient general form in radians is θ = α + 2πn or θ = π − α + 2πn, for integers n.
After obtaining the general families, filter every candidate through the interval specified in the question. The endpoints and the number of whole periods determine how many representatives remain.
For sinθ = 1 or −1, the two families coincide modulo a period, giving one angle per full period rather than two. Recognising this avoids duplicate answers.
A doubled angle changes the number of solutions
Solve 2sin²(2θ) − sin(2θ) − 1 = 0 for 0° ≤ θ < 360°.
Let u = sin(2θ). The quadratic gives u = 1 or −1/2, just as before. But the inner angle φ = 2θ ranges over 0° ≤ φ < 720°.
The condition sinφ = 1 occurs at φ = 90° and 450°. The condition sinφ = −1/2 occurs at φ = 210°, 330°, 570° and 690°.
Divide each φ-value by two. The six θ-solutions are 45°, 105°, 165°, 225°, 285° and 345°. The interval must be transformed with the inner angle; otherwise half the solutions can be missed.
A strict interval changes endpoint inclusion
Suppose a transformed equation yields sinθ = 0. On 0° ≤ θ ≤ 360°, the solutions are 0°, 180° and 360°.
If the interval is 0° < θ < 360°, only 180° remains. If the interval is 0° ≤ θ < 360°, the solutions are 0° and 180°.
The trigonometric equation itself is unchanged; only the allowed angle set differs. A full-circle endpoint pair 0° and 360° represents the same direction, but those are different numerical angles and may both be listed when the interval includes both.
Read the strict or inclusive inequality symbols before constructing the final answer.
Turn the quadratic into a sine inequality
Solve 2sin²θ − sinθ − 1 ≥ 0 on 0° ≤ θ < 360°. Substitute u = sinθ and factorise (2u + 1)(u − 1) ≥ 0.
The quadratic is non-negative when u ≤ −1/2 or u ≥ 1. Intersect this with the sine range [−1,1].
Thus sinθ ≤ −1/2 or sinθ = 1. The first condition gives 210° ≤ θ ≤ 330°, while the second gives θ = 90°.
The answer is the isolated angle 90° together with the closed angular interval from 210° to 330°. This demonstrates why solving an inequality requires sign regions, not only equality roots.
Why sine inequalities can give intervals and isolated angles
For 2u² − u − 1 ≥ 0, the algebraic solution in u is u ≤ −1/2 or u ≥ 1. The admissible range of sine truncates the second branch to the single value u = 1.
The condition u ≤ −1/2 corresponds to a whole arc in the lower half of the unit circle, so it produces an interval of angles. The condition u = 1 corresponds to one maximum point.
This explains the unusual final set: one isolated angle and one interval. It is not a sign-chart mistake.
Students should translate whole u-regions back to angle regions carefully. A solution of an inequality in the substituted variable is not yet an interval for θ until the periodic sine relationship is considered.
A parameter can change whether solutions exist
Consider sin²θ − sinθ + k = 0, with real k and θ allowed to cover one full period. Set u = sinθ.
The equation becomes u² − u + k = 0, or k = u − u². Because u must lie in [−1,1], we ask for the range of g(u) = u − u² on that interval.
Complete the square: g(u) = 1/4 − (u − 1/2)². Its maximum is 1/4 at u = 1/2. At the endpoints, g(−1) = −2 and g(1) = 0, so the minimum on the interval is −2.
Therefore solutions exist exactly when −2 ≤ k ≤ 1/4. The quadratic discriminant alone would give k ≤ 1/4 but would miss the lower restriction created by the sine range.
The parameter endpoint cases are different
At k = 1/4, the substituted quadratic becomes (u − 1/2)² = 0. Its only distinct sine value is u = 1/2, producing two angles in one full period: 30° and 150°.
At k = −2, the substituted equation is u² − u − 2 = (u − 2)(u + 1) = 0. The root u = 2 is inadmissible, while u = −1 gives θ = 270°.
These two endpoint cases both satisfy the existence condition but produce different numbers of distinct angles.
A student should not infer a fixed solution count merely from whether the parameter lies on the boundary of an algebraic discriminant inequality. The sine range and periodic behaviour also matter.
Tangent substitutions have a different inherited range
If a problem is quadratic in tanθ rather than sinθ or cosθ, set u = tanθ. The possible real u-values are unrestricted, but tangent is undefined where cosθ = 0.
For example, tan²θ − 3tanθ + 2 = 0 gives tanθ = 1 or 2. Both are valid tangent outputs.
On 0° ≤ θ < 360°, each nonzero tangent value gives two angles 180° apart, because tangent has period 180°.
The range check is therefore different from a sine substitution. A good learner asks what the new quantity represents instead of copying −1 ≤ u ≤ 1 into every trigonometric quadratic problem.
An equation can be quadratic after an identity
Sometimes the quadratic pattern appears only after replacing a double-angle term. For example, cos2θ = 1 − 2sin²θ.
Consider cos2θ + sinθ = 0. Substitute the identity to obtain 1 − 2sin²θ + sinθ = 0, or 2sin²θ − sinθ − 1 = 0 after multiplying by −1.
This is exactly the opening hidden quadratic. Therefore, on 0° ≤ θ < 360°, the solutions are 90°, 210° and 330°.
The important method choice is to rewrite every term in one trigonometric quantity before factorising. Trying to solve sinθ and cos2θ separately would miss their shared structure.
A check using the original double-angle equation
For θ = 90°, cos2θ = cos180° = −1 and sinθ = 1. Their sum is zero.
For θ = 210°, cos2θ = cos420° = cos60° = 1/2, while sin210° = −1/2. Their sum is zero.
For θ = 330°, cos660° = cos300° = 1/2 and sin330° = −1/2. Again the sum is zero.
These checks confirm that transforming cos2θ through an identity did not change the answer. They also give students a route to catch a sign error in the double-angle identity.
Common mistakes and targeted repairs
A student may solve u correctly but forget that sine and cosine values must stay in [−1,1]. Another may use the right range but list only one angle for sinθ = −1/2.
A third may confuse sin²θ with sin2θ. A fourth may use degree-mode angles while the calculator is set to radians. A fifth may solve the transformed inequality in u but forget to translate an interval into the corresponding unit-circle region.
These are different first errors. Target the missing concept—notation, range, quadrants, periodicity or sign intervals—then assign a changed example without announcing the substitution.
The aim is not merely to memorise the three angles from the opening example, but to recognise the same hidden structure in unfamiliar trigonometric forms.
Practice questions with checked answers
Question one: solve 2sin²θ − sinθ − 1 = 0 on 0° ≤ θ < 360°. Answer: 90°, 210°, 330°.
Question two: solve 2cos²θ + cosθ − 1 = 0 on the same interval. Answer: 60°, 180°, 300°.
Question three: solve sin²θ − 3sinθ + 2 = 0. Answer: 90° only.
Question four: solve cos2θ + sinθ = 0. Answer: 90°, 210°, 330°.
Question five: solve 2sin²θ − sinθ − 1 ≥ 0. Answer: θ = 90° or 210° ≤ θ ≤ 330°.
Question six: find k so that sin²θ − sinθ + k = 0 has a real θ. Answer: −2 ≤ k ≤ 1/4.
For each, write the substituted variable’s range, solve the quadratic and return to every permitted angle.
Syllabus scope and connected learning
Quadratic-in-trigonometric equations connect factorisation, double-angle identities, angle intervals and function ranges. Variations involving parameter existence and solution counts can be useful enrichment, so select exercises according to the learner’s school and examination syllabus.
For Singapore, consult the applicable SEAB syllabus listings for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2.
Related guides include How to List Every Trigonometric Solution on an Interval Crossing Zero, How to Find Parameter Values for a Trigonometric Equation to Have Solutions and How to Spot a Hidden Quadratic in a Logarithmic Equation.
Explore the Additional Mathematics archive or enquire about suitable three-student Mathematics tutorials near Sixth Avenue MRT. The goal is to make the substitution, range check and angle-completeness check three deliberate parts of one solution. Properly taught kids shine a bright light into the future.