Quadratic Trigonometric Equations

Video lesson · from 26:072 worked examples

One lesson video covers all of Trigonometry, so it opens at 26:07 for this topic — not from the beginning.

Theory

Some trigonometric equations take the structure of a quadratic equation. You will need to factorise them to solve.

Sometimes, the equation contains both sine and cosine terms. You must use the identity $\sin^2 x + \cos^2 x = 1$ . Rearrange it to replace one squared term, ensuring the whole equation is expressed using only one trigonometric function.

⚠️ Common Examiner Traps

Factorise — never divide by a trig term: dividing by $\sin x$ throws away every solution where $\sin x = 0$ . Take the common factor out and set each factor to zero.
Substitute to see the quadratic: letting $u = \sin x$ makes the structure obvious and stops sign errors.
Reject impossible values: $\sin x = 2$ has no solutions, since sine never leaves $-1 \leq \sin x \leq 1$ . Say so explicitly rather than ignoring it.
Solve both factors: a quadratic gives two values, and each produces its own set of angles in the domain.

Worked examples

Example 1

Solve $3\sin^2 x^\circ - 4\sin x^\circ + 1 = 0$ for $0 < x^\circ < 360$ .

Let $u = \sin x^\circ$ to see the quadratic structure:

\begin{aligned} 3u^2 - 4u + 1 &= 0 \\ (3u - 1)(u - 1) &= 0 \end{aligned}

This means either $3u - 1 = 0 \implies u = \frac{1}{3}$ or $u - 1 = 0 \implies u = 1$ .

For $\sin x^\circ = \frac{1}{3}$

Base angle $\approx 19.5^\circ$ . Sine is positive in Q1, Q2.

\begin{aligned} x &= 19.5 \\ x &= 180 - 19.5 \\ &= 160.5 \end{aligned}

For $\sin x^\circ = 1$

From the sine graph exact values:

x = 90

Final Solutions: 19.5°, 90°, 160.5°

Example 2

Solve $5\cos^2 x^\circ - 2\cos x^\circ = 3\sin^2 x^\circ$ for $0 < x^\circ < 360$ .

Use $\sin^2 x^\circ = 1 - \cos^2 x^\circ$ to replace the sine term:

\begin{aligned} 5\cos^2 x^\circ - 2\cos x^\circ &= 3(1 - \cos^2 x^\circ) \\ 5\cos^2 x^\circ - 2\cos x^\circ &= 3 - 3\cos^2 x^\circ \end{aligned}

Bring everything to one side to form a quadratic=0:

8\cos^2 x^\circ - 2\cos x^\circ - 3 = 0

Factorise (let $u = \cos x^\circ$ ):

(4\cos x^\circ - 3)(2\cos x^\circ + 1) = 0

For $\cos x^\circ = \frac{3}{4}$

Base angle $\approx 41.4^\circ$ . Cosine is positive in Q1, Q4.

\begin{aligned} x &= 41.4 \\ x &= 360 - 41.4 \\ &= 318.6 \end{aligned}

For $\cos x^\circ = -\frac{1}{2}$

Base angle for $\frac{1}{2}$ is $60^\circ$ . Cosine is negative in Q2, Q3.

\begin{aligned} x &= 180 - 60 \\ &= 120 \\ x &= 180 + 60 \\ &= 240 \end{aligned}

Final Solutions: 41.4°, 120°, 240°, 318.6°