Math Core

An identity like $\sin^2 x + \cos^2 x = 1$ is true for every angle. Most trig equations are different: $2\sin x - 1 = 0$ is true only for certain angles, and your job is to find them. You already know how to find one angle with an inverse trig function. Now you'll find all of them.

Why there is more than one answer

Algebra equations like $2x - 1 = 0$ usually have one solution. Trig equations usually have many, for two reasons.

Symmetry. On the unit circle, two different angles in $[0, 2\pi)$ usually share the same sine (or cosine, or tangent). For example, $\sin \dfrac{\pi}{6}$ and $\sin \dfrac{5\pi}{6}$ both equal $\dfrac{1}{2}$ .
Periodicity. Adding $2\pi$ to an angle lands on the same point of the unit circle, so the trig values repeat forever.

You can see both reasons on a graph. The solutions of $\sin x = \dfrac{1}{2}$ are the $x$ -coordinates where the wave $y = \sin x$ crosses the horizontal line $y = \dfrac{1}{2}$ .

On [0, 2π), the line y = 1/2 meets y = sin x twice: at x = π/6 ≈ 0.52 and x = 5π/6 ≈ 2.62. The pattern repeats every 2π.Open in grapher →

Because of this, a problem always tells you where to look. Most often the interval is $[0, 2\pi)$ in radians or $[0^\circ, 360^\circ)$ in degrees: one full trip around the circle, including $0$ but not $2\pi$ .

The basic method

Solving a basic trig equation

Isolate the trig function, just as you would isolate $x$ : get $\sin x = c$ , $\cos x = c$ or $\tan x = c$ .
Check that a solution exists. $\sin x$ and $\cos x$ are always between $-1$ and $1$ , so $\sin x = 1.4$ has no solution. ( $\tan x$ can equal any number.)
Find the reference angle using $|c|$ : a special angle if you know it, or an inverse trig function if you don't.
Use the sign of $c$ to decide which quadrants the answers are in, and place the reference angle in each one.
List every answer in the interval, or write the general solution by adding multiples of the period.

Recall where each function is positive (the "All Students Take Calculus" pattern): all three in Quadrant I, sine in II, tangent in III, cosine in IV.

Worked example: A sine equation in radians

Solve $2\sin x - 1 = 0$ on $[0, 2\pi)$ .

Isolate: $2\sin x = 1$ , so $\sin x = \dfrac{1}{2}$ .

The reference angle is $\dfrac{\pi}{6}$ , because $\sin \dfrac{\pi}{6} = \dfrac{1}{2}$ .

Sine is positive in Quadrants I and II.

Quadrant I: $x = \dfrac{\pi}{6}$
Quadrant II: $x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$

The solutions are $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$ , matching the two crossings in the graph above.

General solutions

To describe every solution, not only those in one interval, add a whole number of periods. Sine and cosine have period $2\pi$ (or $360^\circ$ ). Tangent has period $\pi$ (or $180^\circ$ ).

For the example above, every solution of $2\sin x - 1 = 0$ is

x = \frac{\pi}{6} + 2\pi k \quad \text{or} \quad x = \frac{5\pi}{6} + 2\pi k, \qquad k \text{ any integer}.

Choosing $k = 1$ gives $\dfrac{13\pi}{6}$ and $\dfrac{17\pi}{6}$ , and $k = -1$ gives $-\dfrac{11\pi}{6}$ and $-\dfrac{7\pi}{6}$ . All of them work.

Worked example: A tangent equation in degrees

Solve $\tan\theta + 1 = 0$ on $[0^\circ, 360^\circ)$ , then write the general solution.

Isolate: $\tan\theta = -1$ . The reference angle is $45^\circ$ .

Tangent is negative in Quadrants II and IV.

Quadrant II: $\theta = 180^\circ - 45^\circ = 135^\circ$
Quadrant IV: $\theta = 360^\circ - 45^\circ = 315^\circ$

Notice that $315^\circ = 135^\circ + 180^\circ$ . Tangent repeats every $180^\circ$ , so one formula covers both:

\theta = 135^\circ + 180^\circ k, \qquad k \text{ any integer}.

When the value isn't special

If $c$ isn't a value you recognize from the unit circle, use an inverse function on your calculator to get the reference angle, then use symmetry exactly as before. Make sure the calculator is in the right mode (radians or degrees).

Worked example: Using inverse cosine

Solve $3\cos x + 1 = 0$ on $[0, 2\pi)$ . Round to the nearest hundredth.

Isolate: $\cos x = -\dfrac{1}{3}$ . Cosine is negative in Quadrants II and III.

A calculator gives $\cos^{-1}\!\left(-\dfrac{1}{3}\right) \approx 1.91$ . That angle is in Quadrant II (between $\dfrac{\pi}{2} \approx 1.57$ and $\pi \approx 3.14$ ), so it is one solution.

For the Quadrant III answer, use the symmetry of cosine: $\cos(2\pi - x) = \cos x$ . So the other solution is

x \approx 2\pi - 1.9106 \approx 4.37.

The solutions are $x \approx 1.91$ and $x \approx 4.37$ .

Common mistake

A calculator's inverse function returns only one angle, and it may not even be in your interval (for example, $\sin^{-1}(-0.5) = -\dfrac{\pi}{6}$ , which is negative). Always ask: which quadrants should the answers be in? Then build every answer from the reference angle.

Equations that factor

Some equations contain the trig function more than once. Treat $\sin x$ (or $\cos x$ ) like a single variable, factor, and set each factor equal to zero. Each factor becomes a basic trig equation.

Worked example: An equation in quadratic form

Solve $2\sin^2 x + \sin x - 1 = 0$ on $[0, 2\pi)$ .

Let $u = \sin x$ . The equation is $2u^2 + u - 1 = 0$ , which factors as $(2u - 1)(u + 1) = 0$ . So

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

$2\sin x - 1 = 0$ gives $\sin x = \dfrac{1}{2}$ , so $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ .
$\sin x + 1 = 0$ gives $\sin x = -1$ , so $x = \dfrac{3\pi}{2}$ (the bottom of the unit circle; only one angle).

The solutions are $\dfrac{\pi}{6}$ , $\dfrac{5\pi}{6}$ and $\dfrac{3\pi}{2}$ .

Tip

Check an answer by substituting it back. For $x = \dfrac{3\pi}{2}$ above: $2(-1)^2 + (-1) - 1 = 2 - 1 - 1 = 0$ . It works.