Trigonometric equations

Solve trig equations using the compound and double-angle formulae: factoring, quadratics in one function, equations with different angles, homogeneous equations, and the danger of squaring.

The strategy

Different angles: factoring

Example 1●●

Solve $\sin 2x = \sin x$ for $0 \le x \le 2\pi$.

$2\sin x\cos x - \sin x = 0$, so $\sin x(2\cos x - 1) = 0$.

$\sin x = 0$: $x = 0, \pi, 2\pi$.

$\cos x = \tfrac12$: $x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$.

Where does y = sin 2x meet y = sin x?

Orange: y = sin 2x. Blue: y = sin x. The five intersections in 0 ≤ x ≤ 2π match the five solutions of Example 1.

Quadratics in one function

Example 2●●

Solve $\cos 2x + \sin x = 0$ for $0^\circ \le x \le 360^\circ$.

$1 - 2\sin^2 x + \sin x = 0$, so $2\sin^2 x - \sin x - 1 = 0$.

Use the form of cos 2x that has only sine in it.

$(2\sin x + 1)(\sin x - 1) = 0$

$\sin x = 1$: $x = 90^\circ$. $\;\sin x = -\tfrac12$: $x = 210^\circ, 330^\circ$.

Homogeneous equations

An equation where every term has the same total degree in $\sin x$ and $\cos x$ is called homogeneous, like $\sin^2 x - 3\sin x\cos x + 2\cos^2 x = 0$ (every term has degree $2$). Divide through by $\cos^2 x$ to get an equation in $\tan x$, after checking that $\cos x = 0$ gives no solutions.

Example 3●●●

Solve $\sin^2 x - 3\sin x\cos x + 2\cos^2 x = 0$ for $0^\circ \le x \le 180^\circ$.

$\tan^2 x - 3\tan x + 2 = 0$

If cos x = 0, the equation says sin²x = 0 too, impossible. So we may divide by cos²x.

$(\tan x - 1)(\tan x - 2) = 0$

$x = 45^\circ$ or $x = \tan^{-1} 2 \approx 63^\circ 26'$.

The danger of squaring

Example 4●●●

Solve $\sin x + \cos x = 1$ for $0^\circ \le x \le 360^\circ$.

Square: $\sin^2 x + 2\sin x\cos x + \cos^2 x = 1$, so $\sin 2x = 0$.

$2x = 0^\circ, 180^\circ, 360^\circ, 540^\circ, 720^\circ$, so $x = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ$.

0 ≤ 2x ≤ 720°.

Checking: $x = 0^\circ, 90^\circ, 360^\circ$.

At 180°: 0 + (−1) = −1. At 270°: −1 + 0 = −1. Both fail.

Flashcards

First step with $sin 2x$ and $sin x$

Expand $\sin 2x = 2\sin x\cos x$, then factor.

Homogeneous equation

Every term has the same degree. Divide by a power of $\cos x$ to get $\tan x$.

After squaring

Check every answer in the original equation.

Practice

How many solutions does $\sin 2x = \sin x$ have for $0 \le x \le 2\pi$?
The first step to solve $\cos 2x + \sin x = 0$ is to replace $\cos 2x$ with:
Solve $\cos 2x + \sin x = 0$ for $0^\circ \le x \le 360^\circ$. Enter the solutions separated by commas.
Why is dividing $\sin 2x = \sin x$ by $\sin x$ a mistake?
Which equation is homogeneous?
Solve $\sin^2 x - 3\sin x\cos x + 2\cos^2 x = 0$ for $0^\circ \le x \le 180^\circ$. Enter the solution that is an exact number of degrees.
Solve $\sin x + \cos x = 1$ for $0^\circ \le x \le 360^\circ$. Enter the solutions separated by commas.
After squaring an equation you must:
Solve $\sin 2x = \cos x$ for $0^\circ \le x \le 360^\circ$. Enter the solutions separated by commas.