Quadratic trigonometric equations

Some trig equations are quadratics in disguise. Others need an identity to get a single trig function first.

Quadratics in sin x or cos x

Treat sin x like a single letter: 2 sin² x - sin x - 1 = 0 factorises as (2 sin x + 1)(sin x - 1) = 0. Solve each bracket separately. Reject values outside -1 to 1.

Using identities

Mixed sin² and cos: replace cos² x with 1 - sin² x (or the other way)
sin x = k cos x: divide by cos x to get tan x = k

Do not divide by sin x

In 2 sin² x = sin x, dividing by sin x loses the solutions where sin x = 0. Rearrange to 0 and factorise: sin x (2 sin x - 1) = 0.

Worked example

Solve 2 sin² x - sin x - 1 = 0 for 0 ≤ x < 360°.

(2 sin x + 1)(sin x - 1) = 0
sin x = -1/2: x = 210° or 330°
sin x = 1: x = 90°

Answer: x = 90°, 210°, 330°

Key idea: Factorise as a quadratic in sin x or cos x, use sin² + cos² = 1 or tan = sin/cos to get one function, and never divide by a trig function.

Check yourself

Solve 2 cos² x - cos x - 1 = 0 for 0 ≤ x < 360°.
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x = 0°, 120° or 240°. (2 cos x + 1)(cos x - 1) = 0. cos x = 1 gives 0°; cos x = -1/2 gives 120° and 240°.
Solve sin x = sqrt(3) cos x for 0 ≤ x < 360°.
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x = 60° or x = 240°. Divide by cos x: tan x = sqrt(3), so x = 60° or 60° + 180° = 240°.
Solve sin² x = sin x for 0 ≤ x < 360°.
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x = 0°, 90° or 180°. sin x (sin x - 1) = 0. sin x = 0 gives 0° and 180°; sin x = 1 gives 90°.
Solve 6 cos² x + sin x - 5 = 0 for 0 ≤ x < 360°. Give the largest solution to 1 decimal place.
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340.5 °. 6(1 - sin² x) + sin x - 5 = 0 gives 6 sin² x - sin x - 1 = 0, so (3 sin x + 1)(2 sin x - 1) = 0. sin x = -1/3 gives x = 199.5° or 340.5°; sin x = 1/2 gives 30° or 150°. The largest is 340.5°.
To solve 2 sin² x = sin x, a student divides both sides by sin x. What goes wrong?
- Nothing: it is a valid method
- It gives extra wrong solutions
- It turns the equation into a cosine equation
- The solutions with sin x = 0 are lost
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The solutions with sin x = 0 are lost. Dividing by sin x assumes sin x is not 0, so x = 0° and 180° are lost.
Solving 3 sin² x + 5 sin x - 2 = 0 gives sin x = 1/3 or sin x = -2. Why is sin x = -2 rejected?
- Negative values of sin x are not allowed
- sin x always lies between -1 and 1
- -2 is not an exact value
- It gives an angle outside the interval
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sin x always lies between -1 and 1. No angle has a sine of -2.
How many solutions does 4 sin² x = 1 have for 0 ≤ x < 360°?
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4. sin x = 1/2 or -1/2. That gives 30°, 150°, 210° and 330°: four solutions.

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Last updated 7 November 2026. Spotted a mistake? Tell us.