Equations solvable by substitution

If an equation contains only $\sin x$ (or only $\cos x$, or only $\tan x$), but with powers or in a non-linear combination, the strategy is to substitute a variable for the trigonometric function, solve the resulting algebraic equation and then go back to the angle. When the equation contains functions of different angles (for example $\cos(2x)$ and $\cos x$), one first uses the trigonometric formulae — such as the double-angle formulae — to reduce everything to a single function.

Example

Solve $2\sin^2 x - 3\sin x + 1 = 0$.

Substitution: $t = \sin x$, with $-1\le t\le 1$. The equation becomes $2t^2 - 3t + 1 = 0 \iff (2t-1)(t-1) = 0 \iff t = \tfrac{1}{2} \ \vee\ t = 1.$ Both solutions satisfy $-1\le t\le 1$, so they are acceptable.

Back to $x$:

• $\sin x = \tfrac{1}{2}$: $\ x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$;
• $\sin x = 1$: $\ x = \dfrac{\pi}{2} + 2k\pi$.

Example

Solve $\cos(2x) + \cos x = 0$.

We apply the double-angle formula $\cos(2x) = 2\cos^2 x - 1$: $2\cos^2 x - 1 + \cos x = 0.$ With $t=\cos x$, $-1\le t\le 1$: $2t^2 + t - 1 = 0 \iff (2t-1)(t+1) = 0 \iff t = \tfrac{1}{2} \ \vee\ t = -1.$

Back to $x$:

• $\cos x = \tfrac{1}{2}$: $\ x = \pm\dfrac{\pi}{3} + 2k\pi$;
• $\cos x = -1$: $\ x = \pi + 2k\pi$.

Links

Topics: Trigonometric equations
Concepts: Trigonometric equation · Double-angle formulae · Substitution
Functions: Cosine · Sine
Methods: Trigonometric equations
Skills: Solving equations · Using formulae