Math Core

What happens when the angle inside the function isn't just $x$ , as in $\sin 2x = \dfrac{\sqrt{3}}{2}$ or $\cos\left(x + \dfrac{\pi}{6}\right) = -\dfrac{1}{2}$ ? These equations show up whenever a wave is stretched or shifted, and a small change in method keeps you from missing solutions.

More cycles, more solutions

Compare $y = \sin x$ and $y = \sin 2x$ on $[0, 2\pi)$ . The graph of $\sin 2x$ has period $\dfrac{2\pi}{2} = \pi$ , so it completes two full cycles in the interval instead of one. A horizontal line that crosses $y = \sin x$ twice will cross $y = \sin 2x$ four times.

y = sin 2x completes two cycles on [0, 2π), so the line y = √3/2 meets it four times.Open in grapher →

This is why the usual method needs one extra step. If you solve only for angles in one trip around the circle, you'll find just half the answers.

The substitution method

The trick is to give the whole inside angle a new name and figure out how far it travels.

Solving an equation with a multiple angle

To solve an equation like $\sin(kx) = c$ on $[0, 2\pi)$ :

Substitute $u = kx$ .
Rescale the interval. If $0 \le x < 2\pi$ , then $0 \le u < 2\pi k$ . The new interval is $k$ times as long.
Solve for $u$ on the new interval: find the answers in $[0, 2\pi)$ , then keep adding $2\pi$ (or $\pi$ for tangent) until you pass the end.
Convert back by dividing each $u$ by $k$ .

The same idea works for shifts: for $u = x + b$ , add $b$ to both ends of the interval.

Worked example: A double angle

Solve $\sin 2x = \dfrac{\sqrt{3}}{2}$ on $[0, 2\pi)$ .

Let $u = 2x$ . Since $0 \le x < 2\pi$ , we have $0 \le u < 4\pi$ .

On $[0, 2\pi)$ , $\sin u = \dfrac{\sqrt{3}}{2}$ when $u = \dfrac{\pi}{3}$ or $u = \dfrac{2\pi}{3}$ . Adding $2\pi$ gives two more that are still less than $4\pi$ :

u = \frac{\pi}{3}, \quad \frac{2\pi}{3}, \quad \frac{7\pi}{3}, \quad \frac{8\pi}{3}.

Divide each by $2$ :

x = \frac{\pi}{6}, \quad \frac{\pi}{3}, \quad \frac{7\pi}{6}, \quad \frac{4\pi}{3}.

These are the four crossings in the graph above.

Common mistake

The most common mistake is dividing by $2$ too early. If you solve $\sin u = \dfrac{\sqrt{3}}{2}$ only on $[0, 2\pi)$ and then divide, you get $\dfrac{\pi}{6}$ and $\dfrac{\pi}{3}$ and lose half the solutions. Stretch the interval first, then solve, then divide.

A quick count

Before you solve, predict how many answers to expect. On $[0, 2\pi)$ , the graph of $\sin kx$ or $\cos kx$ (for a whole number $k$ ) completes $k$ cycles. A typical horizontal line (with $-1 < c < 1$ ) crosses each cycle twice, so you should get $2k$ solutions. For $\tan kx$ , there are $2k$ cycles and one solution per cycle, so again $2k$ solutions. If your list is shorter, you missed some.

Worked example: A triple angle in degrees

Solve $2\cos 3\theta = 1$ on $[0^\circ, 360^\circ)$ .

Isolate: $\cos 3\theta = \dfrac{1}{2}$ . Let $u = 3\theta$ , so $0^\circ \le u < 1080^\circ$ (three full turns).

On the first turn, $\cos u = \dfrac{1}{2}$ at $u = 60^\circ$ and $u = 300^\circ$ . Add $360^\circ$ and $720^\circ$ to each:

u = 60^\circ, \ 300^\circ, \ 420^\circ, \ 660^\circ, \ 780^\circ, \ 1020^\circ.

Divide by $3$ :

\theta = 20^\circ, \ 100^\circ, \ 140^\circ, \ 220^\circ, \ 260^\circ, \ 340^\circ.

That's $2 \cdot 3 = 6$ solutions, as predicted.

Half angles: fewer solutions

When $k$ is a fraction, the interval shrinks. For $\tan \dfrac{x}{2} = 1$ on $[0, 2\pi)$ , let $u = \dfrac{x}{2}$ . Then $0 \le u < \pi$ .

The only angle in $[0, \pi)$ with $\tan u = 1$ is $u = \dfrac{\pi}{4}$ . (The next one, $\dfrac{5\pi}{4}$ , is too big.) Multiply by $2$ : the only solution is $x = \dfrac{\pi}{2}$ .

The wave $y = \tan \dfrac{x}{2}$ is stretched out so much that it only completes one cycle on $[0, 2\pi)$ .

Shifted angles

For an equation like $\sin\left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{2}$ , the angle isn't stretched, it's shifted. Substitute $u = x - \dfrac{\pi}{3}$ and shift the interval by the same amount.

Worked example: A phase shift

Solve $\sin\left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{2}$ on $[0, 2\pi)$ .

Let $u = x - \dfrac{\pi}{3}$ . Subtract $\dfrac{\pi}{3}$ from both ends of $0 \le x < 2\pi$ :

-\frac{\pi}{3} \le u < \frac{5\pi}{3}.

The solutions of $\sin u = \dfrac{1}{2}$ are $\dfrac{\pi}{6} + 2\pi k$ and $\dfrac{5\pi}{6} + 2\pi k$ . The ones in $\left[-\dfrac{\pi}{3}, \dfrac{5\pi}{3}\right)$ are $u = \dfrac{\pi}{6}$ and $u = \dfrac{5\pi}{6}$ . (For example, $\dfrac{\pi}{6} - 2\pi$ is far below $-\dfrac{\pi}{3}$ .)

Add $\dfrac{\pi}{3}$ to convert back:

x = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}, \qquad x = \frac{5\pi}{6} + \frac{\pi}{3} = \frac{7\pi}{6}.

Tip

A graphing check is quick: graph the left side and the right side as two functions and count the intersections on your interval. The number of crossings should match the number of answers on your list.