Harder trig equations come down to the basic ones, using three tools you already have: factoring, identities such as the double-angle formulas, and a careful count of how many turns a multiple angle makes.
Key idea
(1) If it is quadratic in $\sin x$ or $\cos x$, factor it like $au^2 + bu + c$. (2) If a trig function is a common factor, move everything to one side and factor it out; never divide it away. (3) Use identities to get one function of one angle. (4) For $\sin(kx) = c$ on $[0, 2\pi)$, solve for $kx$ on $[0, 2k\pi)$, then divide by $k$.
Why it works
Factoring works because of the zero product property: $AB = 0$ exactly when $A = 0$ or $B = 0$. Dividing both sides by $\sin x$ silently assumes $\sin x \ne 0$, so every solution with $\sin x = 0$ disappears. For multiple angles, as $x$ runs once around, $kx$ runs around $k$ times, picking up $k$ times as many solutions.
Worked examples
Worked example: Quadratic in sine
Solve $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$.
- Let $u = \sin x$: $2u^2 - u - 1 = (2u + 1)(u - 1) = 0$.
- $\sin x = -\frac{1}{2}$: $x = \frac{7\pi}{6}, \frac{11\pi}{6}$. $\sin x = 1$: $x = \frac{\pi}{2}$.
Answer: $\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$
Worked example: A common factor
Solve $2\sin x\cos x = \sin x$ on $[0, 2\pi)$.
- Move everything to one side: $2\sin x\cos x - \sin x = 0$.
- Factor: $\sin x\,(2\cos x - 1) = 0$.
- $\sin x = 0$: $x = 0, \pi$. $\cos x = \frac{1}{2}$: $x = \frac{\pi}{3}, \frac{5\pi}{3}$.
Answer: $0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$
Worked example: A double angle
Solve $\sin 2x = \frac{\sqrt{3}}{2}$ on $[0, 2\pi)$.
- Let $u = 2x$. As $x$ runs over $[0, 2\pi)$, $u$ runs over $[0, 4\pi)$: two turns.
- $\sin u = \frac{\sqrt{3}}{2}$ at $u = \frac{\pi}{3}, \frac{2\pi}{3}$ in the first turn, and $\frac{7\pi}{3}, \frac{8\pi}{3}$ in the second.
- Divide by $2$.
Answer: $\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$
Worked example: Use an identity
Solve $\cos 2x = \cos x$ on $[0, 2\pi)$.
- Use $\cos 2x = 2\cos^2 x - 1$ so only $\cos x$ remains: $2\cos^2 x - \cos x - 1 = 0$.
- Factor: $(2\cos x + 1)(\cos x - 1) = 0$.
- $\cos x = -\frac{1}{2}$: $\frac{2\pi}{3}, \frac{4\pi}{3}$. $\cos x = 1$: $0$.
Answer: $0, \frac{2\pi}{3}, \frac{4\pi}{3}$
Common mistake
Dividing both sides of $2\sin x\cos x = \sin x$ by $\sin x$. You get $\cos x = \frac{1}{2}$ and lose $x = 0$ and $x = \pi$. Factor instead. Check your list against a graph: count the crossings.