Precalculus · F.TF.B.7

Solving basic trig equations

An inverse key hands back one angle with a given sine. How do you find every angle with that sine, and pick out the ones in the stretch you care about?

You need

Inverse trig functionsPrecalculus Exact values on the unit circlePrecalculus

Unlocks

Solving equations with a multiple anglePrecalculus

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

An equation such as \displaystyle \tan x = -\sqrt{3} or 3\cos\theta + 2 = 0 asks for every angle that makes it true. It is evaluation run backward: the unit circle turns an angle into a value, and here the value is given and the angles are wanted.

Two facts shape every answer. A trig function is periodic, so once one angle works, infinitely many do: sine and cosine repeat every 2\pi, tangent every \pi. And within a single turn, sine and cosine usually take each value twice, so even one turn holds two answers. A problem therefore asks for one of two things:

all solutions, an infinite list written as families such as \displaystyle x = \frac{\pi}{6} + 2\pi k, with k any integer;
the solutions in an interval, such as [0, 2\pi), a finite list picked out of those families.

Part 1 (the first day) handles values read straight off the unit circle. Part 2 handles any other value, where an inverse trig function supplies one angle and the symmetry of the circle supplies the second. The same moves find when a Ferris wheel rider reaches a given height or a tide reaches a given level; those lessons put them to work in context.

The procedure

Part 1: get the function alone

In 5\cos x - 1 = 3 the expression \cos x stands for a single unknown number, so the ordinary moves of algebra apply to it: add 1 to both sides, divide by 5, and \displaystyle \cos x = \frac{4}{5}. Every equation in this lesson is first brought to one of the forms \sin x = c, \cos x = c or \tan x = c.

A reciprocal function is turned over. \sec x = -4 says \displaystyle \frac{1}{\cos x} = -4, so \displaystyle \cos x = -\frac{1}{4}. In the same way \csc x = c becomes \displaystyle \sin x = \frac{1}{c}, and \cot x = c becomes \displaystyle \tan x = \frac{1}{c} when c \neq 0.

Part 1: one turn on the unit circle

The angle x ends at the point (\cos x, \sin x) of the unit circle. So

\sin x = c asks where the horizontal line y = c meets the circle;
\cos x = c asks where the vertical line x = c meets it;
\tan x = c asks where the line through the origin with slope c meets it, since \displaystyle \tan x = \frac{\sin x}{\cos x} is the slope of the ray.

A horizontal or vertical line with -1 < c < 1 crosses the circle at two points, so \sin x = c and \cos x = c each have two solutions in one turn. At c = 1 or c = -1 the line only touches the circle, and there is one. When c > 1 or c < -1 it misses, and there are none. A line through the origin crosses the circle at two opposite points, so \tan x = c has two solutions per turn for every c, half a turn apart.

When the size of c is one of the exact values 0, \displaystyle \frac{1}{2}, \displaystyle \frac{\sqrt{2}}{2}, \displaystyle \frac{\sqrt{3}}{2}, 1 (for tangent, \displaystyle \frac{\sqrt{3}}{3}, 1, \displaystyle \sqrt{3}), the points are among the sixteen exact points of the circle. The size of c gives the reference angle, and the sign of c gives the quadrants: sine is positive in Quadrants I and II, cosine in I and IV, tangent in I and III.

Part 1: from one turn to every solution

A full turn brings an angle back to the same point, so once the solutions in one turn are known, every solution is one of them plus 2\pi k, with k any integer. Negative k counts turns backward, so it is needed as much as positive k. For tangent the two crossings are themselves half a turn apart, so one family x_0 + \pi k covers both.

To list the solutions in an interval, take each family, try k = \ldots, -1, 0, 1, 2, \ldots, and keep the angles that land inside. The ends need care: a square bracket includes the end, a round one does not.

Part 2: when the value is not exact

No exact angle has cosine -0.3, but the inverse function supplies one: \arccos(-0.3), the angle in [0, \pi] with that cosine, which a calculator in radian mode gives as a decimal. The inverse returns only the angle in its slice. The circle’s symmetry gives the other.

Figure 1The height c is reached at two points that mirror each other across the y-axis, at \alpha = \arcsin c and at \pi - \alpha.

Sine. The two points at height c mirror each other across the y-axis. With \alpha = \arcsin c, the mirror angle is \pi - \alpha, because \sin(\pi - \alpha) = \sin\alpha.
Cosine. The two points at horizontal coordinate c mirror each other across the x-axis. With \beta = \arccos c, the mirror angle is -\beta, because cosine is even; inside one turn it is 2\pi - \beta.
Tangent. With \gamma = \arctan c, the opposite point is at \gamma + \pi, because tangent has period \pi.

With k any integer, this gives every solution:

\begin{aligned} \sin x = c: \quad x &= \alpha + 2\pi k \\ \text{or} \quad x &= \pi - \alpha + 2\pi k \\ \cos x = c: \quad x &= \pm\beta + 2\pi k \\ \tan x = c: \quad x &= \gamma + \pi k \end{aligned}

Here \alpha = \arcsin c and \beta = \arccos c need -1 \le c \le 1, while \gamma = \arctan c exists for every c. The formulas hold for exact values too; Part 1 simply reads the same angles off the circle.

Worked examples

Example 1

Part 1: a tangent equation, in one turn and in general

Find the solutions of \displaystyle \tan x = -\sqrt{3} in [0, 2\pi), then find all solutions.

Solution

Since \displaystyle \tan\frac{\pi}{3} = \sqrt{3}, the reference angle is \displaystyle \frac{\pi}{3}. Tangent is negative where the coordinates have opposite signs, Quadrants II and IV. The Quadrant II angle with reference angle \displaystyle \frac{\pi}{3} is \displaystyle \pi - \frac{\pi}{3} = \frac{2\pi}{3}, and the Quadrant IV one is \displaystyle 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}. Check: at \displaystyle \frac{2\pi}{3} the point is \displaystyle \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), and \displaystyle \frac{\sqrt{3}}{2} \div \left(-\frac{1}{2}\right) = -\sqrt{3}.

Figure 2Example 1. Every angle with tangent \displaystyle -\sqrt{3} ends on this one line through the origin, at one of its two crossings with the circle. The crossings are half a turn apart.

In [0, 2\pi) the solutions are \displaystyle x = \frac{2\pi}{3} and \displaystyle x = \frac{5\pi}{3}. They are the two ends of one line through the origin, \pi apart, so tangent’s period carries one to the other, and all solutions form a single family:

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

k = 0 gives \displaystyle \frac{2\pi}{3}, k = 1 gives \displaystyle \frac{5\pi}{3}, k = -1 gives \displaystyle -\frac{\pi}{3}, and so on.

Example 2

Part 1: four solutions in two turns

Solve \displaystyle 2\sin x + \sqrt{2} = 0 for all x, then list the solutions in [-2\pi, 2\pi].

Solution

Subtract \displaystyle \sqrt{2} and divide by 2: \displaystyle \sin x = -\frac{\sqrt{2}}{2}. The size \displaystyle \frac{\sqrt{2}}{2} gives the reference angle \displaystyle \frac{\pi}{4}, and sine is negative below the x-axis, in Quadrants III and IV. In one turn that gives \displaystyle \pi + \frac{\pi}{4} = \frac{5\pi}{4} and \displaystyle 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}, so all solutions are

x = \frac{5\pi}{4} + 2\pi k \quad \text{or} \quad x = \frac{7\pi}{4} + 2\pi k

with k any integer. In [-2\pi, 2\pi] the first family gives \displaystyle \frac{5\pi}{4} (k = 0) and \displaystyle -\frac{3\pi}{4} (k = -1); k = 1 gives \displaystyle \frac{13\pi}{4}, too large, and k = -2 gives \displaystyle -\frac{11\pi}{4}, too small. The second family gives \displaystyle \frac{7\pi}{4} and \displaystyle -\frac{\pi}{4}. In order:

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

Two solutions per turn over two turns, as the circle predicts.

Example 3

Part 1: one solution, none, and a reciprocal

Thanh is asked for all solutions of (a) 5\cos\theta + 5 = 0, (b) 3\sin\theta = 4, (c) \sec\theta = 2.

Solution

(a) Subtract 5 and divide by 5: \cos\theta = -1. The vertical line x = -1 touches the circle only at (-1, 0), where \theta = \pi. There is one solution per turn, and all solutions are \theta = \pi + 2\pi k.

(b) \displaystyle \sin\theta = \frac{4}{3}. Every point of the unit circle has height between -1 and 1, and \displaystyle \frac{4}{3} > 1, so there is no solution at all, in any interval.

(c) Turn the secant over: \displaystyle \frac{1}{\cos\theta} = 2, so \displaystyle \cos\theta = \frac{1}{2}. The reference angle is \displaystyle \frac{\pi}{3}, and cosine is positive in Quadrants I and IV, at \displaystyle \frac{\pi}{3} and \displaystyle \frac{5\pi}{3}. Since \displaystyle \frac{5\pi}{3} = -\frac{\pi}{3} + 2\pi, all solutions are \displaystyle \theta = \pm\frac{\pi}{3} + 2\pi k. The cosine is not 0 there, so the secant exists at every one of them.

Example 4

Part 1: degrees over two turns

Hieu needs every \theta with 0^\circ \le \theta < 720^\circ such that \displaystyle 2\cos\theta = -\sqrt{3}.

Solution

Divide by 2: \displaystyle \cos\theta = -\frac{\sqrt{3}}{2}. The size \displaystyle \frac{\sqrt{3}}{2} gives a reference angle of 30^\circ, and cosine is negative in Quadrants II and III: 180^\circ - 30^\circ = 150^\circ and 180^\circ + 30^\circ = 210^\circ. In degrees a full turn is 360^\circ, so the families are 150^\circ + 360^\circ k and 210^\circ + 360^\circ k. The values k = 0 and k = 1 land in the interval; k = 2 is past 720^\circ and k = -1 is negative. The solutions are

\theta = 150^\circ, \ 210^\circ, \ 510^\circ, \ 570^\circ

Example 5

Part 2: an inverse and its mirror

Find all solutions of \cos x = -0.3, and the solutions in [0, 2\pi), to four decimal places.

Solution

-0.3 is not an exact value, so the inverse supplies one angle: \beta = \arccos(-0.3) \approx 1.8755. It lies in [0, \pi], and since \displaystyle \frac{\pi}{2} \approx 1.5708 < 1.8755 < \pi, it is in Quadrant II, as a negative cosine with a positive sine must be. The other point with horizontal coordinate -0.3 is its mirror across the x-axis, in Quadrant III, at -\beta. All solutions are

x \approx \pm 1.8755 + 2\pi k, \quad k \text{ any integer}

In [0, 2\pi) the plus family gives 1.8755 (k = 0). The minus family gives -1.8755 for k = 0, which is negative, and 2\pi - 1.8755 \approx 4.4077 for k = 1. So x \approx 1.8755 or x \approx 4.4077. Check: \cos 4.4077 \approx -0.3000.

Figure 3Example 5. The line y = -0.3 crosses the cosine wave twice in every turn. The large dots are the solutions in [0, 2\pi); every other crossing is one of them shifted by a multiple of 2\pi.

Exactly, the two are \arccos(-0.3) and 2\pi - \arccos(-0.3); the decimals only round them.

Example 6

Part 2: a negative sine and a tangent

Quoc solves each equation on [0, 2\pi), to four decimal places: (a) 4\sin x + 3 = 0; (b) 2\tan x = 5.

Solution

(a) \sin x = -0.75. The inverse gives \alpha = \arcsin(-0.75) \approx -0.8481, a Quadrant IV angle, but negative, so it is not in the interval. Adding a turn gives 2\pi - 0.8481 \approx 5.4351. The mirror angle is \pi - \alpha \approx \pi + 0.8481 \approx 3.9897, in Quadrant III. Both quadrants lie below the x-axis, where sine is negative, as it must be. The solutions are x \approx 3.9897 and x \approx 5.4351.

(b) \tan x = 2.5. The inverse gives \gamma = \arctan 2.5 \approx 1.1903, in Quadrant I. Adding \pi gives 4.3319, in Quadrant III, where tangent is also positive; adding \pi again passes 2\pi. The solutions are x \approx 1.1903 and x \approx 4.3319.

Example 7

Part 2: degrees on an interval centered at zero

Solve 5\cos\theta + 2 = 0 for -360^\circ \le \theta < 360^\circ, to two decimal places.

Solution

Subtract 2 and divide by 5: \displaystyle \cos\theta = -\frac{2}{5}. In degree mode, \displaystyle \beta = \arccos\left(-\frac{2}{5}\right) \approx 113.58^\circ, so all solutions are \theta \approx \pm 113.58^\circ + 360^\circ k. Now take each family through the interval.

k	113.58^\circ + 360^\circ k	-113.58^\circ + 360^\circ k
-1	-246.42^\circ	-473.58^\circ, too small
0	113.58^\circ	-113.58^\circ
1	473.58^\circ, too large	246.42^\circ

The four solutions are \theta \approx -246.42^\circ, -113.58^\circ, 113.58^\circ and 246.42^\circ: two turns, two per turn.

Boundaries

Which values have solutions. \sin x = c and \cos x = c need -1 \le c \le 1; \tan x = c has solutions for every c. So \csc x = c and \sec x = c need c \le -1 or c \ge 1. \cot x = 0 cannot be turned over; it means \cos x = 0, so \displaystyle x = \frac{\pi}{2} + \pi k.
The touching values. When c = 1 or c = -1, sine and cosine have one solution per turn, and the two families of the formulas merge into one: for c = 1, \displaystyle \alpha = \frac{\pi}{2} and \displaystyle \pi - \alpha = \frac{\pi}{2}.
All solutions or an interval. “All solutions” is an infinite list, and k runs over every integer, negative ones included. “Solutions in an interval” is a finite list, and it changes with the ends: [0, 2\pi] and [0, 2\pi) differ whenever 0 is a solution.
Units. Radians unless degrees are asked for; in degrees, 2\pi becomes 360^\circ and \pi becomes 180^\circ.
Rounding. A decimal solution is an approximation of an exact one such as \arccos(-0.3). Round only at the end.
One function of x alone. The method needs a single sine, cosine or tangent with x by itself inside. A multiple inside, as in \sin 2x, or two functions or a square in the same equation, needs the further work of the next lessons.

Where this goes next

When the angle inside is a multiple such as 3x, solving equations with a multiple angle solves for 3x exactly as here and then divides, and the extra cycles bring extra solutions. Factoring and identities break a harder equation into several basic ones, each finished by this lesson. Trigonometric inequalities begin by solving the matching equation: its solutions are the points where the function crosses the level, and the answer is the stretches between them.

Exercises

Part 1: exact solutions

Find all solutions of \displaystyle \sin x = \frac{\sqrt{3}}{2}.
Find the solutions of 4\cos x + 2 = 0 in [0, 2\pi).
Find all solutions of \tan x + 1 = 0.
Find the solutions of \displaystyle \sqrt{3}\tan x = 1 in [0, 2\pi).
Find all solutions of 3\sin x - 3 = 0.
Find all solutions of \cos x = 0, as a single family.
Find the solutions of \displaystyle \sec x = -\sqrt{2} in [0, 2\pi).
Find all solutions of 2\cos x = 3.

Part 1: other intervals

Find the solutions of \displaystyle 2\sin x = \sqrt{3} in [-2\pi, 2\pi].
Find the solutions of \displaystyle \tan x = \sqrt{3} in [0, 4\pi).
Find every \theta with -360^\circ \le \theta < 360^\circ such that \displaystyle \cos\theta = \frac{\sqrt{2}}{2}.
How many solutions does \displaystyle \cos x = \frac{1}{2} have in [0, 10\pi)?

Part 2: with the inverse functions

Give radians to four decimal places and degrees to two.

Find the solutions of \sin x = 0.35 in [0, 2\pi).
Find all solutions of \cos x = 0.6.
Find the solutions of 5\tan x + 7 = 0 in [0, 2\pi).
Find the solutions of 2\cos x + 1.6 = 0 in [0, 2\pi).
Find all solutions of \csc x = 3.
Find every \theta with 0^\circ \le \theta < 360^\circ such that 8\sin\theta + 5 = 0.
Find the solutions of \cot x = 0.5 in [0, 2\pi).
Find the solutions of \sin x = -0.9 in [\pi, 3\pi).

Reasoning

Explain why \tan x = c has exactly one solution in every interval \displaystyle \left(-\frac{\pi}{2} + \pi m, \frac{\pi}{2} + \pi m\right), for any integer m, while \sin x = c with -1 < c < 1 has two in every full turn.
Thanh solves \sin x = 0.2, finds \arcsin 0.2 \approx 0.2014, and writes “all solutions are x \approx 0.2014 + 2\pi k”. What is missing? Give the complete answer.
Explain, with the unit circle, why every solution of \cos x = c is \beta + 2\pi k or -\beta + 2\pi k, where \beta = \arccos c, and why 2\pi - \beta belongs to the second family.
For which values of c does \sin x = c have exactly one solution in [0, 2\pi)? Explain.
Quoc claims that for every number c, the equations \sin x = c and \cos x = c have the same number of solutions in [0, 2\pi). Is he right? Explain.

Answers

\displaystyle x = \frac{\pi}{3} + 2\pi k or \displaystyle x = \frac{2\pi}{3} + 2\pi k. Reference angle \displaystyle \frac{\pi}{3}; sine is positive in Quadrants I and II.
\displaystyle x = \frac{2\pi}{3} or \displaystyle x = \frac{4\pi}{3}. \displaystyle \cos x = -\frac{1}{2}, reference angle \displaystyle \frac{\pi}{3}, cosine negative in Quadrants II and III.
\displaystyle x = \frac{3\pi}{4} + \pi k. \tan x = -1, reference angle \displaystyle \frac{\pi}{4}; the Quadrant II angle is \displaystyle \frac{3\pi}{4}, and adding \pi reaches the Quadrant IV one.
\displaystyle x = \frac{\pi}{6} or \displaystyle x = \frac{7\pi}{6}. \displaystyle \tan x = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}, reference angle \displaystyle \frac{\pi}{6}, tangent positive in Quadrants I and III.
\displaystyle x = \frac{\pi}{2} + 2\pi k. \sin x = 1; the line y = 1 touches the circle only at the top, (0, 1).
\displaystyle x = \frac{\pi}{2} + \pi k. The line x = 0 meets the circle at (0, 1) and (0, -1), at \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}, which are \pi apart, so one family with step \pi covers both.
\displaystyle x = \frac{3\pi}{4} or \displaystyle x = \frac{5\pi}{4}. \displaystyle \cos x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}, reference angle \displaystyle \frac{\pi}{4}, cosine negative in Quadrants II and III.
No solution. \displaystyle \cos x = \frac{3}{2}, and no point of the unit circle has horizontal coordinate above 1.
\displaystyle x = -\frac{5\pi}{3}, -\frac{4\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}. \displaystyle \sin x = \frac{\sqrt{3}}{2} gives the families \displaystyle \frac{\pi}{3} + 2\pi k and \displaystyle \frac{2\pi}{3} + 2\pi k; use k = 0 and k = -1.
\displaystyle x = \frac{\pi}{3}, \frac{4\pi}{3}, \frac{7\pi}{3}, \frac{10\pi}{3}. The family \displaystyle \frac{\pi}{3} + \pi k for k = 0, 1, 2, 3.
\theta = -315^\circ, -45^\circ, 45^\circ, 315^\circ. The families \pm 45^\circ + 360^\circ k; 45^\circ - 360^\circ = -315^\circ and -45^\circ + 360^\circ = 315^\circ.
10 solutions. In each of the five turns [2\pi m, 2\pi m + 2\pi), m = 0, 1, 2, 3, 4, there are two, \displaystyle \frac{\pi}{3} + 2\pi m and \displaystyle \frac{5\pi}{3} + 2\pi m.
x \approx 0.3576 or x \approx 2.7840. \arcsin 0.35 \approx 0.3576, and the mirror angle is \pi - \arcsin 0.35.
x \approx \pm 0.9273 + 2\pi k. \arccos 0.6 \approx 0.9273, and cosine is even.
x \approx 2.1910 or x \approx 5.3326. \tan x = -1.4 and \arctan(-1.4) \approx -0.9505; add \pi and 2\pi to the unrounded angle.
x \approx 2.4981 or x \approx 3.7851. \cos x = -0.8, \arccos(-0.8) \approx 2.4981, and 2\pi - \arccos(-0.8) \approx 3.7851.
x \approx 0.3398 + 2\pi k or x \approx 2.8018 + 2\pi k. \displaystyle \sin x = \frac{1}{3}, \displaystyle \arcsin\frac{1}{3} \approx 0.3398, and \pi - 0.3398 \approx 2.8018.
\theta \approx 218.68^\circ or \theta \approx 321.32^\circ. \sin\theta = -0.625 and \arcsin(-0.625) \approx -38.68^\circ; adding 360^\circ gives 321.32^\circ, and the mirror is 180^\circ + 38.68^\circ = 218.68^\circ.
x \approx 1.1071 or x \approx 4.2487. \displaystyle \tan x = \frac{1}{0.5} = 2, \arctan 2 \approx 1.1071, and adding \pi gives 4.2487.
x \approx 4.2614 or x \approx 5.1634. \arcsin(-0.9) \approx -1.1198; the mirror is \pi + 1.1198 \approx 4.2614, and -1.1198 + 2\pi \approx 5.1634. Both lie in [\pi, 3\pi), and the next ones, 4.2614 + 2\pi and 5.1634 - 2\pi, do not.
On each interval \displaystyle \left(-\frac{\pi}{2} + \pi m, \frac{\pi}{2} + \pi m\right) tangent rises from below every number to above every number, without breaks, so it passes each value c once: the tangent graph has one rising branch there, between two asymptotes. Sine over a full turn rises from -1 to 1 and falls back; a height strictly between -1 and 1 is passed once on the way up and once on the way down, so twice. On the circle: the line y = c cuts the circle twice, while each line through the origin meets the open right or left half of the circle once.
The mirror family is missing. Sine also equals 0.2 at \pi - 0.2014 \approx 2.9402, in Quadrant II. All solutions: x \approx 0.2014 + 2\pi k or x \approx 2.9402 + 2\pi k.
A solution is an angle whose terminal point has horizontal coordinate c. The vertical line x = c meets the circle at \displaystyle (c, \sqrt{1 - c^2}) and \displaystyle (c, -\sqrt{1 - c^2}), mirror images across the x-axis. The first is reached by \beta, since \beta lies in [0, \pi] and ends on the upper half; reflecting across the x-axis turns an angle into its negative, so the second is reached by -\beta. Every angle ending at one of the two points differs from \beta or -\beta by whole turns, 2\pi k. Finally 2\pi - \beta = -\beta + 2\pi \cdot 1, the second family with k = 1.
c = 1 and c = -1. The line y = c must meet the circle exactly once, which happens only when it touches the circle at the top or bottom: \displaystyle x = \frac{\pi}{2} or \displaystyle x = \frac{3\pi}{2}. For -1 < c < 1 (including c = 0, with solutions 0 and \pi) there are two, and otherwise none.
He is right. For -1 < c < 1 a horizontal line and a vertical line each cross the circle twice, so both equations have two solutions. For c = \pm 1 each line touches the circle once: \sin x = 1 at \displaystyle \frac{\pi}{2}, \cos x = 1 at 0, \sin x = -1 at \displaystyle \frac{3\pi}{2}, \cos x = -1 at \pi. For c > 1 or c < -1 neither has a solution.