Trig Equations with Multiple Angles: Steps

Trig equations with multiple angles have the variable multiplied inside the function, like $\sin 2x = \frac{\sqrt{3}}{2}$ . To solve one, let $u = 2x$ , stretch the interval to match, and find every value of $u$ . Then divide each one by 2. A multiple angle usually gives more solutions in $[0, 2\pi)$ .

Before this lesson: You should know Solving Trig Equations, The Unit Circle and Amplitude and Period.

Why a Multiple Angle Gives More Solutions

The graph of $y = \sin 2x$ has period $\pi$ , not $2\pi$ . So it fits two full waves into $[0, 2\pi)$ . Each wave can cross a horizontal line twice, so $\sin 2x = \frac{\sqrt{3}}{2}$ has four solutions there.

The same idea works for any number $k$ . The graph of $\sin kx$ repeats $k$ times in $[0, 2\pi)$ . So you can expect up to $2k$ solutions, two for each wave.

Graph with x from 0 to 2 pi and y from −1.5 to 1.5. The blue curve y = sin 2x makes two full waves. It crosses the orange dashed line y = √3/2 at four points, x = pi/6, pi/3, 7 pi/6 and 4 pi/3, which are marked with green dots and gray dashed lines down to the x-axis. — The graph of y = sin 2x makes two waves on [0, 2π), so it meets y = √3/2 four times.

Steps for Trig Equations with Multiple Angles

To solve an equation like $\sin kx = c$ on $[0, 2\pi)$ , follow these steps.

Isolate the trig function.
Let $u = kx$ . Multiply the ends of the interval by $k$ . If $0 \le x \lt 2\pi$ , then $0 \le u \lt 2k\pi$ .
Solve for $u$ in $[0, 2\pi)$ with reference angles, just like in solving trig equations.
Add $2\pi$ to each answer again and again, as long as it stays below $2k\pi$ . For tangent, add $\pi$ instead.
Divide every value of $u$ by $k$ to get $x$ .

For the general solution, write $u$ with $+ 2\pi n$ , then divide everything by $k$ . For example, $2x = \frac{\pi}{3} + 2\pi n$ becomes $x = \frac{\pi}{6} + \pi n$ .

Worked Examples

Example 1: Sine of 2x

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

0 \le u \lt 4\pi

Let $u = 2x$ . Multiply the interval by 2.

The reference angle is $\frac{\pi}{3}$ , because $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

Sine is positive in Quadrants I and II: $u = \frac{\pi}{3}$ and $u = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

u = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}

Add $2\pi$ to the first.

u = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}

Add $2\pi$ to the second.

Adding $2\pi$ again gives $\frac{13\pi}{3}$ , which is more than $4\pi$ , so stop

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

Divide each $u$ by 2.

Answer: $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$ . These are the four crossing points in the graph above.

Example 2: Cosine of 3x

Solve $2 \cos 3x + 1 = 0$ on $[0, 2\pi)$ .

2 \cos 3x = -1

Subtract 1 from both sides.

\cos 3x = -\frac{1}{2}

Divide both sides by 2.

0 \le u \lt 6\pi

Let $u = 3x$ . Multiply the interval by 3.

The reference angle is $\frac{\pi}{3}$ . Cosine is negative in Quadrants II and III: $u = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $u = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

Add $2\pi$ to each: $u = \frac{8\pi}{3}$ and $u = \frac{10\pi}{3}$

Add $2\pi$ again: $u = \frac{14\pi}{3}$ and $u = \frac{16\pi}{3}$ . Both are less than $6\pi = \frac{18\pi}{3}$

x = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}, \frac{10\pi}{9}, \frac{14\pi}{9}, \frac{16\pi}{9}

Divide each $u$ by 3.

Answer: six solutions, $x = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}, \frac{10\pi}{9}, \frac{14\pi}{9}, \frac{16\pi}{9}$ . Three waves give two crossings each.

Example 3: A half angle

Solve $\tan \frac{x}{2} = 1$ on $[0, 2\pi)$ , and give the general solution.

0 \le u \lt \pi

Let $u = \frac{x}{2}$ . Multiply the interval by $\frac{1}{2}$ .

The reference angle is $\frac{\pi}{4}$ , because $\tan \frac{\pi}{4} = 1$

Tangent is positive in Quadrants I and III, but $\pi + \frac{\pi}{4} = \frac{5\pi}{4}$ is not less than $\pi$ . So the only value is $u = \frac{\pi}{4}$

x = \frac{\pi}{2}

Multiply $u$ by 2.

\frac{x}{2} = \frac{\pi}{4} + \pi n

General solution for $u$ .

x = \frac{\pi}{2} + 2\pi n

Multiply both sides by 2.

Answer: $x = \frac{\pi}{2}$ on $[0, 2\pi)$ . A fraction inside makes the graph stretch out, so there are fewer solutions, not more.

Example 4: A multiple angle with a shift

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

\sin(2x - \frac{\pi}{3}) = \frac{1}{2}

Divide both sides by 2.

0 \le 2x \lt 4\pi

Let $u = 2x - \frac{\pi}{3}$ . Multiply the interval by 2.

-\frac{\pi}{3} \le u \lt \frac{11\pi}{3}

Subtract $\frac{\pi}{3}$ from each part.

Sine is $\frac{1}{2}$ at $u = \frac{\pi}{6}$ and $u = \frac{5\pi}{6}$

Add $2\pi$ to each: $u = \frac{13\pi}{6}$ and $u = \frac{17\pi}{6}$ . Both are less than $\frac{11\pi}{3} = \frac{22\pi}{6}$

Check below the interval: $\frac{5\pi}{6} - 2\pi = -\frac{7\pi}{6}$ , which is less than $-\frac{\pi}{3}$ , so it does not count

2x = \frac{3\pi}{6}, \frac{7\pi}{6}, \frac{15\pi}{6}, \frac{19\pi}{6}

Add $\frac{\pi}{3} = \frac{2\pi}{6}$ to each $u$ (see adding and subtracting fractions)

2x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{5\pi}{2}, \frac{19\pi}{6}

Simplify.

x = \frac{\pi}{4}, \frac{7\pi}{12}, \frac{5\pi}{4}, \frac{19\pi}{12}

Divide each by 2.

Answer: $x = \frac{\pi}{4}, \frac{7\pi}{12}, \frac{5\pi}{4}, \frac{19\pi}{12}$ . The shift moves the interval for $u$ , which is covered in phase shift.

Example 5: Word problem with a spring

A weight bounces on a spring. Its position is $y = 5 \cos(2\pi t)$ centimeters from rest, $t$ seconds after release. During the first 2 seconds, $0 \le t \lt 2$ , when is the weight 2.5 cm above rest?

5 \cos(2\pi t) = 2.5

Set the position to 2.5.

\cos(2\pi t) = \frac{1}{2}

Divide both sides by 5.

0 \le u \lt 4\pi

Let $u = 2\pi t$ . Multiply the interval by $2\pi$ .

Cosine is positive in Quadrants I and IV: $u = \frac{\pi}{3}$ and $u = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Add $2\pi$ to each: $u = \frac{7\pi}{3}$ and $u = \frac{11\pi}{3}$

t = \frac{1}{6}, \frac{5}{6}, \frac{7}{6}, \frac{11}{6}

Divide each $u$ by $2\pi$ .

t \approx 0.17, 0.83, 1.17, 1.83

Write as decimals.

Answer: at about 0.17, 0.83, 1.17 and 1.83 seconds. The spring bounces once per second, so it passes 2.5 cm twice each second. Models like this come from sinusoidal models.

Common Mistakes

Mistake	Example of the error	Fix
Not stretching the interval	For $\sin 2x = \frac{\sqrt{3}}{2}$ , giving only $x = \frac{\pi}{6}, \frac{\pi}{3}$	Use $0 \le 2x \lt 4\pi$ and add $2\pi$ to find all four.
Dividing too early	Writing $\sin x = \frac{\sqrt{3}}{4}$ from $\sin 2x = \frac{\sqrt{3}}{2}$	The 2 is inside the function. Solve for $2x$ first.
Treating $\sin 2x$ as $2 \sin x$	Rewriting $\sin 2x = 1$ as $2 \sin x = 1$	They are different. Use $u = 2x$ , or the double-angle formulas.
Dividing only the first part of the period	$2x = \frac{\pi}{3} + 2\pi n$ , so $x = \frac{\pi}{6} + 2\pi n$	Divide every term by 2: $x = \frac{\pi}{6} + \pi n$ .
Keeping values out of range	For $\tan \frac{x}{2} = 1$ , keeping $u = \frac{5\pi}{4}$	Here $u$ must be less than $\pi$ , so only $u = \frac{\pi}{4}$ works.

Related Methods

Find the general solution first

You can also solve for all $u$ at once, then divide. For $\sin 2x = \frac{\sqrt{3}}{2}$ , write $2x = \frac{\pi}{3} + 2\pi n$ or $2x = \frac{2\pi}{3} + 2\pi n$ . Divide by 2: $x = \frac{\pi}{6} + \pi n$ or $x = \frac{\pi}{3} + \pi n$ . Then plug in $n = 0, 1, 2, \ldots$ and keep the values below $2\pi$ .

Count the waves first

The period of $\sin kx$ or $\cos kx$ is $\frac{2\pi}{k}$ , so it makes $k$ waves in $[0, 2\pi)$ . Most values give 2 solutions per wave. This tells you how many answers to look for before you start.

Check with a graph

Graph each side as its own function and count where they cross, as in the picture above. Use radian mode, and make sure the window shows all of $[0, 2\pi)$ .

Practice Problems

Solve $\sin 2x = 1$ on $[0, 2\pi)$ .
Solve $\cos 2x = \frac{1}{2}$ on $[0, 2\pi)$ .
Solve $\tan 2x = 1$ on $[0, 2\pi)$ .
Solve $2 \sin 3x = \sqrt{2}$ on $[0, 2\pi)$ .
Solve $\sin \frac{x}{2} = \frac{1}{2}$ on $[0, 2\pi)$ .
Solve $\sin 4x = 0$ on $[0, 2\pi)$ .
Solve $2 \cos 2x + \sqrt{3} = 0$ on $[0, 2\pi)$ .
Write the general solution of $\sin 3x = -1$ .
Solve $\cos(2x + \frac{\pi}{3}) = \frac{1}{2}$ on $[0, 2\pi)$ .
A wave machine makes the water height $h = 0.5 \sin(\pi t) + 2$ meters, $t$ seconds after it starts. During the first 4 seconds, $0 \le t \lt 4$ , when is the height 2.25 meters?

Show answers

Let $u = 2x$ , so $0 \le u \lt 4\pi$
Sine is 1 at $u = \frac{\pi}{2}$

u = \frac{5\pi}{2}

Add $2\pi$ .

x = \frac{\pi}{4}, \frac{5\pi}{4}

Divide by 2.

Answer: $x = \frac{\pi}{4}, \frac{5\pi}{4}$

Let $u = 2x$ , so $0 \le u \lt 4\pi$
Cosine is positive in Quadrants I and IV: $u = \frac{\pi}{3}$ and $u = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

u = \frac{7\pi}{3}, \frac{11\pi}{3}

Add $2\pi$ to each.

x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}

Divide by 2.

Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

Let $u = 2x$ , so $0 \le u \lt 4\pi$
Tangent is 1 at $u = \frac{\pi}{4}$ , and tangent repeats every $\pi$

u = \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}

Add $\pi$ three times.

x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}

Divide by 2.

Answer: $x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$

\sin 3x = \frac{\sqrt{2}}{2}

Divide both sides by 2.

Let $u = 3x$ , so $0 \le u \lt 6\pi$
Sine is positive in Quadrants I and II: $u = \frac{\pi}{4}$ and $u = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

u = \frac{9\pi}{4}, \frac{11\pi}{4}

Add $2\pi$ .

u = \frac{17\pi}{4}, \frac{19\pi}{4}

Add $2\pi$ again.

x = \frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}

Divide by 3.

Answer: the six values above

Let $u = \frac{x}{2}$ , so $0 \le u \lt \pi$
Sine is positive in Quadrants I and II: $u = \frac{\pi}{6}$ and $u = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
Both are less than $\pi$ , and adding $2\pi$ goes past $\pi$

x = \frac{\pi}{3}, \frac{5\pi}{3}

Multiply by 2.

Answer: $x = \frac{\pi}{3}, \frac{5\pi}{3}$

Let $u = 4x$ , so $0 \le u \lt 8\pi$

u = 0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi, 7\pi

Sine is 0 at every multiple of $\pi$ .

x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}

Divide by 4.

Answer: those 8 values

2 \cos 2x = -\sqrt{3}

Subtract $\sqrt{3}$ from both sides.

\cos 2x = -\frac{\sqrt{3}}{2}

Divide both sides by 2.

Let $u = 2x$ , so $0 \le u \lt 4\pi$
Cosine is negative in Quadrants II and III: $u = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ and $u = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$

u = \frac{17\pi}{6}, \frac{19\pi}{6}

Add $2\pi$ .

x = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}

Divide by 2.

Answer: $x = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}$

3x = \frac{3\pi}{2} + 2\pi n

Sine is $-1$ only at $\frac{3\pi}{2}$ in one turn.

x = \frac{\pi}{2} + \frac{2\pi n}{3}

Divide every term by 3.

Answer: $x = \frac{\pi}{2} + \frac{2\pi n}{3}$ , where $n$ is any integer

0 \le 2x \lt 4\pi

Let $u = 2x + \frac{\pi}{3}$ . Multiply the interval by 2.

\frac{\pi}{3} \le u \lt \frac{13\pi}{3}

Add $\frac{\pi}{3}$ .

Cosine is $\frac{1}{2}$ at $u = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$ ( $\frac{13\pi}{3}$ is not included)

2x = 0, \frac{4\pi}{3}, 2\pi, \frac{10\pi}{3}

Subtract $\frac{\pi}{3}$ .

x = 0, \frac{2\pi}{3}, \pi, \frac{5\pi}{3}

Divide by 2.

Answer: $x = 0, \frac{2\pi}{3}, \pi, \frac{5\pi}{3}$

10.

0.5 \sin(\pi t) + 2 = 2.25

Set the height to 2.25.

0.5 \sin(\pi t) = 0.25

Subtract 2 from both sides.

\sin(\pi t) = \frac{1}{2}

Divide both sides by 0.5.

Let $u = \pi t$ , so $0 \le u \lt 4\pi$

u = \frac{13\pi}{6}, \frac{17\pi}{6}

$u = \frac{\pi}{6}, \frac{5\pi}{6}$ , then add $2\pi$ .

t = \frac{1}{6}, \frac{5}{6}, \frac{13}{6}, \frac{17}{6}

Divide by $\pi$ .

Answer: at about 0.17, 0.83, 2.17 and 2.83 seconds

FAQ

How do you solve trig equations with multiple angles?

Replace the angle with $u$ , multiply the interval by the same number, and find every value of $u$ in that bigger interval. Then divide each value by the number to get $x$ .

Why does sin 2x have more solutions than sin x?

The graph of $\sin 2x$ repeats twice as fast. It makes two waves in $[0, 2\pi)$ instead of one, so it usually crosses a horizontal line four times instead of two.

Is sin 2x the same as 2 sin x?

No. For example, at $x = \frac{\pi}{2}$ , $\sin 2x = \sin \pi = 0$ , but $2 \sin x = 2$ . The double-angle formula says $\sin 2x = 2 \sin x \cos x$ .

How do you find the general solution with a multiple angle?

Write the general solution for the whole angle, then divide every term by the number in front of $x$ . For $3x = \frac{\pi}{2} + 2\pi n$ , you get $x = \frac{\pi}{6} + \frac{2\pi n}{3}$ .

Summary

Let $u = kx$ and multiply the interval by $k$ .
Find every $u$ in the new interval by adding $2\pi$ (or $\pi$ for tangent).
Divide each value of $u$ by $k$ to get $x$ .
A whole number $k$ gives more solutions. A fraction gives fewer.
For the general solution, divide every term, including $2\pi n$ , by $k$ .

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