Math Core

An identity is true for every input. A trigonometric equation is true only for certain inputs, and your job is to find them all. Because trig functions are periodic, a solvable equation usually has infinitely many solutions, so the answer is either a list on one period, such as $[0, 2\pi)$ , or a formula that captures every solution at once.

The basic equation

Every trig equation eventually comes down to one of the form $\sin x = c$ , $\cos x = c$ or $\tan x = c$ . Solve it in two stages:

Find all solutions in one period. For sine and cosine that period is $[0, 2\pi)$ , and there are usually two solutions, one in each quadrant where the function has the sign of $c$ . Tangent has period $\pi$ , so one solution per $\pi$ .
Add multiples of the period to get every solution.

On [0, 2π) the line y = −1/2 meets y = sin x at 7π/6 ≈ 3.67 and 11π/6 ≈ 5.76.Open in grapher →

Worked example: A linear equation

Solve $2\sin x + 1 = 0$ . Give the solutions on $[0, 2\pi)$ and the general solution.

Solution. Isolate the trig function: $\sin x = -\dfrac{1}{2}$ . The reference angle is $\dfrac{\pi}{6}$ , and sine is negative in Quadrants III and IV, so on $[0, 2\pi)$

x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \qquad\text{or}\qquad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}.

Every solution is one of these plus a whole number of full turns:

x = \frac{7\pi}{6} + 2\pi k \quad\text{or}\quad x = \frac{11\pi}{6} + 2\pi k, \qquad k \text{ an integer}.

When $c$ is not a special value, use the inverse function for one solution and symmetry for the other. For $\sin x = c$ , the solutions on one period are $\sin^{-1} c$ and $\pi - \sin^{-1} c$ . For $\cos x = c$ , they are $\pm\cos^{-1} c$ . For $\tan x = c$ , it's $\tan^{-1} c + \pi k$ .

General solutions

For an integer $k$ :

\begin{aligned} \sin x = c &\iff x = \sin^{-1} c + 2\pi k \ \text{ or } \ x = \pi - \sin^{-1} c + 2\pi k \quad (-1 \le c \le 1) \\ \cos x = c &\iff x = \pm\cos^{-1} c + 2\pi k \quad (-1 \le c \le 1) \\ \tan x = c &\iff x = \tan^{-1} c + \pi k \quad (\text{any } c) \end{aligned}

If $|c| > 1$ , then $\sin x = c$ and $\cos x = c$ have no solutions.

Equations of quadratic type

If the same trig function appears squared and to the first power, treat it like a variable. Setting $u = \cos x$ turns $2\cos^2 x - \cos x - 1 = 0$ into $2u^2 - u - 1 = 0$ . Factor or use the quadratic formula, then solve each basic equation. Throw out any value of $u$ outside the range of the function.

When the equation mixes two functions, use a Pythagorean identity to rewrite it in just one.

Worked example: Use an identity, then factor

Solve $2\cos^2 x + 3\sin x = 3$ on $[0, 2\pi)$ .

Solution. Replace $\cos^2 x$ with $1 - \sin^2 x$ so only sine remains:

2(1 - \sin^2 x) + 3\sin x - 3 = 0 \quad\Longrightarrow\quad 2\sin^2 x - 3\sin x + 1 = 0.

Factor: $(2\sin x - 1)(\sin x - 1) = 0$ .

$\sin x = \dfrac{1}{2}$ gives $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ .
$\sin x = 1$ gives $x = \dfrac{\pi}{2}$ .

The solutions are $\dfrac{\pi}{6}$ , $\dfrac{\pi}{2}$ and $\dfrac{5\pi}{6}$ .

Factor, don't divide

When the same trig function appears in every term, it's tempting to divide it out. Don't. Dividing by $\sin x$ silently throws away every solution where $\sin x = 0$ .

Worked example: A double angle

Solve $\sin 2x = \cos x$ on $[0, 2\pi)$ .

Solution. Rewrite with a single angle: $2\sin x\cos x = \cos x$ . Move everything to one side and factor:

2\sin x\cos x - \cos x = 0 \quad\Longrightarrow\quad \cos x\,(2\sin x - 1) = 0.

$\cos x = 0$ gives $x = \dfrac{\pi}{2}$ or $x = \dfrac{3\pi}{2}$ .
$\sin x = \dfrac{1}{2}$ gives $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ .

So $x = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$ . Dividing by $\cos x$ at the start would have lost two of the four.

Common mistake

Never divide both sides by an expression that can equal zero, such as $\sin x$ , $\cos x$ or $\tan x$ . Factor it out instead and set each factor equal to zero.

Multiple angles

For an equation like $\sin 3x = c$ , solve for the whole angle $\theta = 3x$ first, then divide. The catch is the interval. If $0 \le x < 2\pi$ , then $0 \le 3x < 6\pi$ , which is three full turns, so you should expect three times as many solutions.

Worked example: Solving for a multiple angle

Solve $2\cos 2x = \sqrt3$ on $[0, 2\pi)$ .

Solution. $\cos 2x = \dfrac{\sqrt3}{2}$ . Let $\theta = 2x$ , which runs over $[0, 4\pi)$ . Cosine equals $\dfrac{\sqrt3}{2}$ at $\theta = \dfrac{\pi}{6}$ and $\dfrac{11\pi}{6}$ in the first turn, and one full turn later at $\dfrac{13\pi}{6}$ and $\dfrac{23\pi}{6}$ .

Divide each by $2$ :

x = \frac{\pi}{12}, \ \frac{11\pi}{12}, \ \frac{13\pi}{12}, \ \frac{23\pi}{12}.

Equivalently, from the general solution $2x = \pm\dfrac{\pi}{6} + 2\pi k$ you get $x = \pm\dfrac{\pi}{12} + \pi k$ : the period has been cut in half.

Squaring can add false solutions

Sometimes the only way to get to one function is to square both sides, for example in $\sin x + \cos x = 1$ . Squaring gives $1 + 2\sin x\cos x = 1$ , so $\sin 2x = 0$ , which suggests $x = 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}$ . But squaring also accepts solutions of $\sin x + \cos x = -1$ . Check each candidate in the original equation: $x = \pi$ gives $0 + (-1) = -1$ and $x = \dfrac{3\pi}{2}$ gives $-1 + 0 = -1$ , so both are extraneous. The true solutions are $0$ and $\dfrac{\pi}{2}$ .

Tip

A graph is a fast check on how many solutions to expect. Graph each side as its own function over one period and count the intersections. If you found three solutions but the curves cross four times, go back and look for the missing one.

A good order of attack for any trig equation: rewrite in a single angle, rewrite in a single function if possible, collect everything on one side, factor, solve each basic equation, and check if you squared.