When the angle inside is or instead of , how many solutions are there, and how do you find every one?

Precalculus · F.TF.B.7

Solving equations with a multiple angle

When the angle inside is 3x or \displaystyle \frac{x}{3} instead of x, how many solutions are there, and how do you find every one?

You need

Solving basic trig equationsPrecalculus

Where you are 4 in Solving trigonometric equations

Trigonometry7

Trigonometric identities13

Solving trigonometric equations4

Solving basic trig equations Trigonometric inequalities

Solving equations with a multiple angle

Solving trig equations by factoring, quadratics and identities

When you reach for it

Solving a basic trig equation finds every x with \sin x = c, \cos x = c or \tan x = c. Often the unknown sits inside the function multiplied by a number: \displaystyle 2\sin(3x) = \sqrt{2}, \displaystyle \cos\left(\frac{x}{3}\right) = -\frac{1}{2}, \tan(2x) = -1. Such equations come from waves whose period is not 2\pi, and they are what a model such as a Ferris wheel or a tide produces when you ask when it reaches a given level.

The method is the basic one applied to the inner angle, with one new fact: on a fixed interval of x, the inner angle covers a stretched interval. For \sin(3x) it runs through three turns while x runs through one, so there are three times as many solutions. For \displaystyle \cos\left(\frac{x}{3}\right) it runs through a third of a turn, and there may be fewer solutions, or none.

The procedure

Name the inner angle

In \sin(3x) = c, the angle the sine acts on is 3x, not x. Call it u = 3x. Then the equation reads \sin u = c, a basic equation, and everything known about basic equations applies to u.

This is allowed because u = bx with b \neq 0 matches each x with exactly one u and each u with exactly one \displaystyle x = \frac{u}{b}. So x solves \sin(bx) = c exactly when u = bx solves \sin u = c: finding all the u and dividing each by b finds all the x, no more and no fewer.

Stretch the interval along with the angle

When x is restricted, u is restricted too. For b > 0, multiplying each part of 0 \le x < 2\pi by b keeps the inequalities pointing the same way:

0 \le x < 2\pi \quad \Longleftrightarrow \quad 0 \le bx < 2\pi b

So u runs over [0, 2\pi b), which is b turns. For b = 3 that is [0, 6\pi), three turns, and since \sin u = c with -1 < c < 1 has two solutions per turn, \sin(3x) = c has six solutions in [0, 2\pi). For \displaystyle b = \frac{1}{3} the inner angle covers only \displaystyle \left[0, \frac{2\pi}{3}\right), a third of a turn, so it holds at most two of the solutions, possibly none.

The same move works on any interval: p \le x < q becomes bp \le u < bq, with the same kind of ends.

Divide at the end, step included

All solutions of \sin u = c are u = \alpha + 2\pi k or u = \pi - \alpha + 2\pi k, with \alpha = \arcsin c and k any integer. Dividing by b divides the whole expression, the step 2\pi k included:

\begin{aligned} x &= \frac{\alpha}{b} + \frac{2\pi}{b}k \\ \text{or} \quad x &= \frac{\pi - \alpha}{b} + \frac{2\pi}{b}k \end{aligned}

The new step \displaystyle \frac{2\pi}{b} is exactly the period of \sin(bx), as the period from the b-value shows: the solutions repeat once per cycle of the graph. For \cos(bx) = c the families are \displaystyle x = \pm\frac{\beta}{b} + \frac{2\pi}{b}k with \beta = \arccos c, and for \tan(bx) = c the single family is \displaystyle x = \frac{\gamma}{b} + \frac{\pi}{b}k with \gamma = \arctan c.

There are two ways to reach the solutions in an interval, and they always agree. List the solutions for u in the stretched interval, then divide each by b; or divide the families first, then try k = 0, 1, 2, \ldots and k = -1, -2, \ldots until the values leave the interval of x. The worked examples use both.

Worked examples

Example 1

Six solutions from three turns

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

Solution

Divide by 2: \displaystyle \sin(3x) = \frac{\sqrt{2}}{2}. Let u = 3x. As x runs over [0, 2\pi), u runs over [0, 6\pi), three full turns.

In the first turn, \displaystyle \sin u = \frac{\sqrt{2}}{2} has reference angle \displaystyle \frac{\pi}{4}, and sine is positive in Quadrants I and II, so \displaystyle u = \frac{\pi}{4} or \displaystyle u = \frac{3\pi}{4}. Each further turn adds \displaystyle 2\pi = \frac{8\pi}{4}:

\begin{aligned} u = {} & \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{9\pi}{4}, \\ & \frac{11\pi}{4}, \ \frac{17\pi}{4}, \ \frac{19\pi}{4} \end{aligned}

The next would be \displaystyle \frac{25\pi}{4}, past \displaystyle 6\pi = \frac{24\pi}{4}. Now \displaystyle x = \frac{u}{3}:

\begin{aligned} x = {} & \frac{\pi}{12}, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \\ & \frac{11\pi}{12}, \ \frac{17\pi}{12}, \ \frac{19\pi}{12} \end{aligned}

Check one: \displaystyle 3 \cdot \frac{11\pi}{12} = \frac{11\pi}{4} = \frac{3\pi}{4} + 2\pi, whose sine is \displaystyle \frac{\sqrt{2}}{2}, so \displaystyle 2\sin(3x) = \sqrt{2}.

Figure 1Example 1. Three full waves of y = 2\sin(3x) fit in [0, 2\pi), and the line \displaystyle y = \sqrt{2} crosses each wave twice, so there are six solutions.

Stopping after the first turn would have found only \displaystyle \frac{\pi}{12} and \displaystyle \frac{\pi}{4}, a third of the answer.

Example 2

Fewer cycles, fewer solutions

Find all solutions of \displaystyle \cos\left(\frac{x}{3}\right) = -\frac{1}{2}, then the solutions in [0, 2\pi].

Solution

Let \displaystyle u = \frac{x}{3}, so \displaystyle \cos u = -\frac{1}{2}. The reference angle is \displaystyle \frac{\pi}{3}, and cosine is negative in Quadrants II and III, at \displaystyle \frac{2\pi}{3} and \displaystyle \frac{4\pi}{3} = -\frac{2\pi}{3} + 2\pi. So \displaystyle u = \pm\frac{2\pi}{3} + 2\pi k, and multiplying by 3 (dividing by \displaystyle \frac{1}{3}) gives every solution:

x = \pm 2\pi + 6\pi k, \quad k \text{ any integer}

The step 6\pi is the period of \displaystyle \cos\left(\frac{x}{3}\right).

For 0 \le x \le 2\pi, the inner angle runs over \displaystyle 0 \le u \le \frac{2\pi}{3}, a third of a turn. Of all the solutions for u, only \displaystyle \frac{2\pi}{3} lies there, at the very end, so the one solution is

x = 2\pi

Check: \displaystyle \cos\frac{2\pi}{3} = -\frac{1}{2}. Had the interval been [0, 2\pi), that end would be left out, and there would be no solution at all.

Figure 2Example 2. One wave of \displaystyle y = \cos\left(\frac{x}{3}\right) takes 6\pi. Over the thick stretch [0, 2\pi] only a third of the wave lies, and the line \displaystyle y = -\frac{1}{2} is reached once, at the very end.

Example 3

A tangent with a doubled angle

Find all solutions of \tan(2x) = -1, then the solutions in [0, 2\pi).

Solution

Let u = 2x. Since \displaystyle \tan\frac{\pi}{4} = 1, the reference angle is \displaystyle \frac{\pi}{4}; tangent is negative in Quadrant II, at \displaystyle \frac{3\pi}{4}, and tangent’s period \pi reaches the Quadrant IV angle. So \displaystyle u = \frac{3\pi}{4} + \pi k, and dividing everything by 2:

x = \frac{3\pi}{8} + \frac{\pi}{2}k, \quad k \text{ any integer}

Now try values of k. k = 0, 1, 2, 3 give \displaystyle \frac{3\pi}{8}, \displaystyle \frac{7\pi}{8}, \displaystyle \frac{11\pi}{8}, \displaystyle \frac{15\pi}{8}; k = 4 gives \displaystyle \frac{19\pi}{8}, past \displaystyle 2\pi = \frac{16\pi}{8}, and k = -1 gives \displaystyle -\frac{\pi}{8}, below 0. The solutions in [0, 2\pi) are

x = \frac{3\pi}{8}, \ \frac{7\pi}{8}, \ \frac{11\pi}{8}, \ \frac{15\pi}{8}

The count fits the stretch: u covers [0, 4\pi), four half turns, one solution in each. None of these lands where \tan(2x) is undefined, at \displaystyle 2x = \frac{\pi}{2} + \pi k, the asymptotes of the graph.

Example 4

An inverse, then divide

Solve 5\cos(2x) - 2 = 0 on [0, 2\pi), to four decimal places.

Solution

Add 2 and divide by 5: \cos(2x) = 0.4. Let u = 2x, which runs over [0, 4\pi), two turns. The inverse gives \beta = \arccos 0.4 \approx 1.1593; keep the calculator’s full value. In the first turn the solutions are \beta and its mirror 2\pi - \beta; the second turn adds 2\pi to each, giving 2\pi + \beta and 4\pi - \beta. Dividing by 2:

\begin{aligned} x &= \frac{\beta}{2}, \quad \pi - \frac{\beta}{2}, \\ x &= \pi + \frac{\beta}{2}, \quad 2\pi - \frac{\beta}{2} \end{aligned}

To four decimal places, x \approx 0.5796, 2.5620, 3.7212, 5.7035. Check: \cos(2 \cdot 3.7212) = \cos 7.4424 \approx 0.400.

The exact values are what get divided; rounding \beta first and then halving can shift the last digit.

Example 5

Counting without listing

Ngoc wants only the number of solutions in [0, 2\pi) of (a) \sin(5x) = 0.3 and (b) \cos(4x) = 1.

Solution

(a) With u = 5x, the inner angle covers [0, 10\pi), which splits into five turns [2\pi m, 2\pi m + 2\pi) for m = 0, 1, 2, 3, 4. Since -1 < 0.3 < 1, the line at height 0.3 crosses the unit circle twice, so each turn holds two solutions. There are 10.

(b) With u = 4x, the inner angle covers [0, 8\pi), four turns. But \cos u = 1 only touches the circle, at (1, 0), once per turn: u = 0, 2\pi, 4\pi, 6\pi, while 8\pi is excluded by the round bracket. There are 4, namely \displaystyle x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}.

The count “two per turn, b turns” needs both a value strictly between -1 and 1 and a whole number of turns.

Example 6

When \pi is the multiplier

Huong solves 2\sin(\pi t) + 1 = 0 for 0 \le t < 4.

Solution

Subtract 1 and divide by 2: \displaystyle \sin(\pi t) = -\frac{1}{2}. Let u = \pi t. Multiplying 0 \le t < 4 by \pi gives 0 \le u < 4\pi, two turns. Sine is negative in Quadrants III and IV, and the reference angle is \displaystyle \frac{\pi}{6}, so the first turn gives \displaystyle \frac{7\pi}{6} and \displaystyle \frac{11\pi}{6}, and the second adds \displaystyle 2\pi = \frac{12\pi}{6}:

u = \frac{7\pi}{6}, \ \frac{11\pi}{6}, \ \frac{19\pi}{6}, \ \frac{23\pi}{6}

Dividing by \pi removes the \pi:

t = \frac{7}{6}, \ \frac{11}{6}, \ \frac{19}{6}, \ \frac{23}{6}

The period of \sin(\pi t) is \displaystyle \frac{2\pi}{\pi} = 2, and indeed \displaystyle \frac{19}{6} = \frac{7}{6} + 2 and \displaystyle \frac{23}{6} = \frac{11}{6} + 2. Models in time, where one cycle takes a number of seconds or hours, give answers like these with no \pi in them.

Boundaries

b \neq 0. With b = 0 the function is a constant, such as \sin 0 = 0, and there is no inner angle to solve for.
A negative multiplier. For b < 0, multiplying an inequality by b reverses it, so the stretched interval turns around. It is simpler to rewrite first: \sin(-3x) = -\sin(3x) and \tan(-3x) = -\tan(3x) because sine and tangent are odd, and \cos(-3x) = \cos(3x) because cosine is even.
The count 2b. “Twice b solutions in [0, 2\pi)” holds only for sine or cosine, a positive integer b and -1 < c < 1. At c = \pm 1 it is b; for any other b, as in Example 2, list and count.
The ends of the interval. The stretched interval keeps the brackets of the original. A solution of u at an end, as in Example 2, is kept or dropped by the same bracket.
Which values have solutions. As for basic equations, \sin(bx) = c and \cos(bx) = c need -1 \le c \le 1, and \tan(bx) = c has solutions for every c. Multiplying the angle changes how many solutions fit in an interval, never which values are reached.
A constant added inside. In \displaystyle \sin(2x - \frac{\pi}{3}) = c the inner angle is \displaystyle u = 2x - \frac{\pi}{3}, the same angle that factoring before reading a phase shift writes as \displaystyle 2\left(x - \frac{\pi}{6}\right); the interval for u is found the same way (multiply, then subtract), and each x comes from \displaystyle x = \frac{u + \pi/3}{2}.
Rounding. Divide exact values, and round only the final x.

Where this goes next

Factoring and identities turn harder equations into several basic or multiple-angle ones; a factor such as \sin(2x) = 0 is finished by this lesson. Trigonometric inequalities start from the crossing points of a wave and a level, and for \sin(bx) those crossings are the solutions found here. Every sinusoidal model, from a Ferris wheel to a tide, asks for a multiple-angle equation as soon as it asks “when”.

Exercises

Exact solutions in [0, 2\pi)

\displaystyle 2\cos(2x) = -\sqrt{3}
2\sin(4x) = 1
\tan(3x) = 1
\displaystyle \sin\left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}
\cos(2x) = 0
\sin(3x) = -1

All solutions

Find all solutions of \displaystyle 2\sin(2x) = -\sqrt{3}.
Find all solutions of \displaystyle \tan\left(\frac{x}{2}\right) = 1.
Find all solutions of \cos(5x) = 1.
Find all solutions of \sec(3x) = 2.

Other intervals and counts

Solve \displaystyle 2\cos\left(\frac{x}{2}\right) + \sqrt{2} = 0 for 0 \le x < 4\pi.
Solve \displaystyle \sin(\pi x) = \frac{\sqrt{2}}{2} for 0 \le x < 4.
Find every \theta with 0^\circ \le \theta < 180^\circ such that \displaystyle \cos(3\theta) = -\frac{1}{2}.
How many solutions does \sin(6x) = 0.8 have in [0, 2\pi)?
How many solutions does \displaystyle \cos\left(\frac{x}{4}\right) = \frac{1}{2} have in [0, 2\pi)? Find them.

With the inverse functions

Give answers to four decimal places.

Solve \sin(2x) = 0.6 on [0, 2\pi).
Solve 4\cos(3x) + 1 = 0 on [0, 2\pi).
Solve \displaystyle \tan\left(\frac{x}{2}\right) = 4 on [0, 2\pi).
Find all solutions of \cos(2x) = -0.7.

Reasoning

Vinh solves \displaystyle \sin(2x) = \frac{1}{2} on [0, 2\pi), finds \displaystyle 2x = \frac{\pi}{6} or \displaystyle 2x = \frac{5\pi}{6}, divides, and answers \displaystyle \frac{\pi}{12} and \displaystyle \frac{5\pi}{12}. What did he miss? Give the full answer.
Ngoc claims that for every number c, \sin(2x) = c has exactly twice as many solutions in [0, 2\pi) as \sin x = c. Is she right? Explain.
Explain why the solutions of \cos(bx) = c, for b > 0, repeat every \displaystyle \frac{2\pi}{b}, and how this matches the graph of y = \cos(bx).
Huong looks at \displaystyle \cos\left(\frac{x}{3}\right) = c on [0, 2\pi). For which c is there no solution, and for which exactly one? Explain.
Vinh writes all solutions of \cos(3x) = 0 as \displaystyle x = \frac{\pi}{6} + \pi k. Find his error and give the correct family.

Answers

\displaystyle x = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}. \displaystyle \cos u = -\frac{\sqrt{3}}{2} with u = 2x in [0, 4\pi): \displaystyle u = \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{17\pi}{6}, \frac{19\pi}{6}; halve.
\displaystyle x = \frac{\pi}{24}, \displaystyle \frac{5\pi}{24}, \displaystyle \frac{13\pi}{24}, \displaystyle \frac{17\pi}{24}, \displaystyle \frac{25\pi}{24}, \displaystyle \frac{29\pi}{24}, \displaystyle \frac{37\pi}{24}, \displaystyle \frac{41\pi}{24}. \displaystyle \sin u = \frac{1}{2} with u = 4x in [0, 8\pi): \displaystyle \frac{\pi}{6}, \frac{5\pi}{6} plus 2\pi, 4\pi, 6\pi; divide by 4.
\displaystyle x = \frac{\pi}{12}, \displaystyle \frac{5\pi}{12}, \displaystyle \frac{3\pi}{4}, \displaystyle \frac{13\pi}{12}, \displaystyle \frac{17\pi}{12}, \displaystyle \frac{7\pi}{4}. \displaystyle u = 3x = \frac{\pi}{4} + \pi k in [0, 6\pi) for k = 0 to 5, so \displaystyle x = \frac{\pi}{12} + \frac{\pi}{3}k.
\displaystyle x = \frac{2\pi}{3}, \frac{4\pi}{3}. \displaystyle u = \frac{x}{2} runs over [0, \pi), where \displaystyle \sin u = \frac{\sqrt{3}}{2} at \displaystyle \frac{\pi}{3} and \displaystyle \frac{2\pi}{3}; double.
\displaystyle x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}. \displaystyle u = 2x = \frac{\pi}{2} + \pi k in [0, 4\pi) for k = 0 to 3.
\displaystyle x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}. \displaystyle u = 3x = \frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, one per turn of [0, 6\pi).
\displaystyle x = \frac{2\pi}{3} + \pi k or \displaystyle x = \frac{5\pi}{6} + \pi k. \displaystyle \sin u = -\frac{\sqrt{3}}{2} gives \displaystyle u = \frac{4\pi}{3} + 2\pi k or \displaystyle \frac{5\pi}{3} + 2\pi k; halve, step included.
\displaystyle x = \frac{\pi}{2} + 2\pi k. \displaystyle \frac{x}{2} = \frac{\pi}{4} + \pi k; double.
\displaystyle x = \frac{2\pi}{5}k. 5x = 2\pi k, since cosine equals 1 only at whole turns.
\displaystyle x = \pm\frac{\pi}{9} + \frac{2\pi}{3}k. \displaystyle \cos(3x) = \frac{1}{2}, so \displaystyle 3x = \pm\frac{\pi}{3} + 2\pi k; divide by 3.
\displaystyle x = \frac{3\pi}{2}, \frac{5\pi}{2}. \displaystyle \cos u = -\frac{\sqrt{2}}{2} with \displaystyle u = \frac{x}{2} in [0, 2\pi): \displaystyle u = \frac{3\pi}{4}, \frac{5\pi}{4}; double.
\displaystyle x = \frac{1}{4}, \frac{3}{4}, \frac{9}{4}, \frac{11}{4}. u = \pi x in [0, 4\pi): \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}; divide by \pi.
\theta = 40^\circ, 80^\circ, 160^\circ. u = 3\theta in [0^\circ, 540^\circ) with \displaystyle \cos u = -\frac{1}{2}: 120^\circ, 240^\circ, 480^\circ (600^\circ is too large).
12. u = 6x covers six turns, two solutions each since -1 < 0.8 < 1.
One, \displaystyle x = \frac{4\pi}{3}. \displaystyle u = \frac{x}{4} covers \displaystyle \left[0, \frac{\pi}{2}\right), which holds only \displaystyle u = \frac{\pi}{3}.
x \approx 0.3218, 1.2490, 3.4633, 4.3906. With \alpha = \arcsin 0.6, u = 2x is \alpha, \pi - \alpha, 2\pi + \alpha, 3\pi - \alpha; halve.
x \approx 0.6078, 1.4866, 2.7022, 3.5810, 4.7966, 5.6754. \cos u = -0.25, \beta = \arccos(-0.25); u = 3x is \beta, 2\pi - \beta, 2\pi + \beta, 4\pi - \beta, 4\pi + \beta, 6\pi - \beta; divide by 3.
x \approx 2.6516. \displaystyle u = \frac{x}{2} covers [0, \pi), holding only u = \arctan 4; double.
x \approx \pm 1.1731 + \pi k. 2x = \pm\arccos(-0.7) + 2\pi k; halve, step included.
He stopped after one turn of u = 2x, which runs over [0, 4\pi). Adding 2\pi gives \displaystyle 2x = \frac{13\pi}{6}, \frac{17\pi}{6}. The full answer is \displaystyle x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}.
She is right. With u = 2x, the inner angle covers [0, 4\pi), the two turns [0, 2\pi) and [2\pi, 4\pi). Each holds exactly as many solutions of \sin u = c as [0, 2\pi) does, since one is the other shifted by 2\pi. So the count doubles for every c: 2 becomes 4, 1 becomes 2 at c = \pm 1, and 0 stays 0.
If \cos(bx) = c, then \displaystyle b\left(x + \frac{2\pi}{b}\right) = bx + 2\pi, one full turn past bx, so \displaystyle \cos\left(b\left(x + \frac{2\pi}{b}\right)\right) = c as well. So adding \displaystyle \frac{2\pi}{b} to a solution gives another, and subtracting works the same way. On the graph, \displaystyle \frac{2\pi}{b} is the period of y = \cos(bx): a horizontal line meets each cycle in the same places.
No solution for \displaystyle c \le -\frac{1}{2} or c > 1; exactly one for \displaystyle -\frac{1}{2} < c \le 1. The inner angle \displaystyle u = \frac{x}{3} covers \displaystyle \left[0, \frac{2\pi}{3}\right), where cosine falls steadily from 1 toward \displaystyle -\frac{1}{2} without reaching it. So each value in \displaystyle \left(-\frac{1}{2}, 1\right] is taken once, and no other value is taken.
He divided only the starting angle, not the step. \displaystyle 3x = \frac{\pi}{2} + \pi k gives \displaystyle x = \frac{\pi}{6} + \frac{\pi}{3}k. His family misses solutions such as \displaystyle \frac{\pi}{2}, where \displaystyle \cos\frac{3\pi}{2} = 0.