Unit 6, lesson 53

Solving Basic Trigonometric Equations

How to find every solution of a basic sine, cosine or tangent equation in a given interval, including scaled inputs and quadratics in a trigonometric function, and how to check the list.

  • Theory part 105
  • Practice part 106
  • Identities and Equations

A Ferris wheel car passes a given height many times during a long ride, twice on each turn unless that height is the very top or bottom. A trigonometric equation behaves the same way. It usually has infinitely many solutions, so the real work is to find all of them that lie in the interval you care about, and to be sure none are missing. In calculus this is how you locate the critical points of a function like sin⁡2x\sin 2x on a closed interval.

Key terms

Trigonometric equation

Definition A trigonometric equation is an equation in which the unknown appears inside one or more trigonometric functions, for example sin⁡x=c\sin x = c or 2cos⁡2x−cos⁡x=12\cos^2 x - \cos x = 1. Angles are measured in radians. A solution is a real number xx at which every function in the equation is defined and the two sides are equal. To solve the equation means to find the set of all solutions, either on the whole real line or in an interval stated with the problem.

In plain words You are looking for the angles that make the equation true. There are usually many of them, because the trigonometric functions repeat.

Example The equation sin⁡x=12\sin x = \frac{1}{2} is true at x=π6x = \frac{\pi}{6} and at x=5π6x = \frac{5\pi}{6}, and also at every angle that differs from one of these by a full turn, such as π6+2π=13π6\frac{\pi}{6} + 2\pi = \frac{13\pi}{6}. The equation sin⁡x=2\sin x = 2 has no solutions at all, since −1≤sin⁡x≤1-1 \le \sin x \le 1 for every real xx.

Reference angle

Definition Let θ\theta be an angle in standard position whose terminal side does not lie on a coordinate axis. Its reference angle is the acute angle θref\theta_{\text{ref}}, with 0<θref<π20 < \theta_{\text{ref}} < \frac{\pi}{2}, between the terminal side and the xx-axis. For θ\theta in [0,2π)[0, 2\pi) it is

θref={θin quadrant I,π−θin quadrant II,θ−πin quadrant III,2π−θin quadrant IV.\theta_{\text{ref}} = \begin{cases} \theta & \text{in quadrant I}, \\ \pi - \theta & \text{in quadrant II}, \\ \theta - \pi & \text{in quadrant III}, \\ 2\pi - \theta & \text{in quadrant IV}. \end{cases}

The values of sine, cosine and tangent at θ\theta equal their values at θref\theta_{\text{ref}} up to sign, and the quadrant decides the sign.

In plain words The reference angle tells you how far the angle sits from the horizontal axis. The size of the trigonometric value comes from it, and the sign comes from the quadrant.

Example To solve sin⁡u=−32\sin u = -\frac{\sqrt{3}}{2} on [0,2π)[0, 2\pi), start from sin⁡π3=32\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, so the reference angle is π3\frac{\pi}{3}. Sine is negative in quadrants III and IV, which gives

u=π+π3=4π3andu=2π−π3=5π3.\begin{gathered} u = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \quad \text{and} \\[0.4em] u = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}. \end{gathered}

Unit circle intersections

Definition The point at angle uu on the unit circle is (cos⁡u,sin⁡u)(\cos u, \sin u). So the solutions of sin⁡u=c\sin u = c are the angles of the points where the horizontal line y=cy = c meets the circle, and the solutions of cos⁡u=c\cos u = c are the angles of the points where the vertical line x=cx = c meets it. For −1<c<1-1 < c < 1 each line meets the circle in exactly two points, for c=±1c = \pm 1 in exactly one point, and for ∣c∣>1\lvert c \rvert > 1 in none. The solutions of tan⁡u=c\tan u = c are the angles of the two points where the line y=cxy = cx through the origin meets the circle, and these two points are opposite each other.

In plain words Draw the line, see where it crosses the circle, and read off the angles. The picture tells you how many solutions to expect in one turn before you compute anything.

Example For cos⁡u=−12\cos u = -\frac{1}{2} the vertical line x=−12x = -\frac{1}{2} crosses the circle once above the axis and once below. The reference angle is π3\frac{\pi}{3}, so in [0,2π)[0, 2\pi) the two crossing angles are 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}, and there are no others.

Periodic solutions

Definition Sine and cosine have period 2π2\pi and tangent has period π\pi. If α\alpha is any one solution, then every real solution is given as follows, where kk is an integer.

sin⁡u=sin⁡α⟺u=α+2kπ  or u=π−α+2kπ,cos⁡u=cos⁡α⟺u=α+2kπ  or u=−α+2kπ,tan⁡u=tan⁡α⟺u=α+kπ.\begin{aligned} &\sin u = \sin \alpha \quad \Longleftrightarrow \\[0.4em] &\quad u = \alpha + 2k\pi \ \text{ or } \\[0.4em] &\quad u = \pi - \alpha + 2k\pi, \\[0.4em] &\cos u = \cos \alpha \quad \Longleftrightarrow \\[0.4em] &\quad u = \alpha + 2k\pi \ \text{ or } \\[0.4em] &\quad u = -\alpha + 2k\pi, \\[0.4em] &\tan u = \tan \alpha \quad \Longleftrightarrow \\[0.4em] &\quad u = \alpha + k\pi. \end{aligned}

In the tangent case, α\alpha and uu must both avoid the odd multiples of π2\frac{\pi}{2}, where tangent is undefined. A formula of this kind is called the general solution.

In plain words Find the solutions in one full period, then add every whole number of periods to each of them.

Example Taking α=4π3\alpha = \frac{4\pi}{3} in the sine formula gives π−α=−π3\pi - \alpha = -\frac{\pi}{3}, which is the same position as 5π3\frac{5\pi}{3}. So every real solution of sin⁡u=−32\sin u = -\frac{\sqrt{3}}{2} is

u=4π3+2kπoru=5π3+2kπ,k∈Z.\begin{gathered} u = \frac{4\pi}{3} + 2k\pi \quad \text{or} \\[0.4em] u = \frac{5\pi}{3} + 2k\pi, \\[0.4em] k \in \mathbb{Z}. \end{gathered}

Integer parameter

Definition The letter kk in a general solution is an integer parameter. It ranges over all of Z={…,−2,−1,0,1,2,… }\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}, negative values included, and each integer gives one solution from each family. To find the solutions in a bounded interval, solve the interval inequality for kk and keep exactly the integers that satisfy it.

In plain words Each value of kk moves you around the circle a whole number of times. You try the integers that land you inside the interval and ignore the rest.

Example For the family u=4π3+2kπu = \frac{4\pi}{3} + 2k\pi on [−2π,2π][-2\pi, 2\pi] we need

−2π≤4π3+2kπ≤2π⟺−53≤k≤13,\begin{gathered} -2\pi \le \frac{4\pi}{3} + 2k\pi \le 2\pi \\[0.4em] \Longleftrightarrow \quad -\frac{5}{3} \le k \le \frac{1}{3}, \end{gathered}

so k=−1k = -1 or k=0k = 0, which gives u=−2π3u = -\frac{2\pi}{3} and u=4π3u = \frac{4\pi}{3}. The same work for u=5π3+2kπu = \frac{5\pi}{3} + 2k\pi gives u=−π3u = -\frac{\pi}{3} and u=5π3u = \frac{5\pi}{3}. So the equation sin⁡u=−32\sin u = -\frac{\sqrt{3}}{2} has exactly four solutions in [−2π,2π][-2\pi, 2\pi].

Specified interval

Definition When a problem asks for the solutions in an interval II, the answer is the set of all solutions of the equation that belong to II. State II first, then choose the periodic representatives that fall in it. Every such solution must be listed, and an endpoint of II is included exactly when the interval is closed at that end and the endpoint solves the equation. If no restriction is given, the answer is the general solution.

In plain words The interval is part of the question. Whether its ends are brackets or parentheses can change the answer.

Example On [0,2π][0, 2\pi] the equation sin⁡x=0\sin x = 0 has the three solutions 00, π\pi and 2π2\pi. On [0,2π)[0, 2\pi) it has only 00 and π\pi, because the right endpoint is left out.

Scaled input

Definition An equation such as sin⁡(Bx+C)=c\sin(Bx + C) = c with constants B≠0B \ne 0 and CC has a scaled input. Put u=Bx+Cu = Bx + C. If xx runs over [a,b][a, b], then uu runs over

[Ba+C, Bb+C]  when B>0,[Bb+C, Ba+C]  when B<0.\begin{gathered} [Ba + C, \, Bb + C] \ \text{ when } B > 0, \\[0.4em] [Bb + C, \, Ba + C] \ \text{ when } B < 0. \end{gathered}

A negative BB reverses the inequalities, so the endpoints swap. Solve the equation for uu in the new interval, then solve back for x=u−CBx = \frac{u - C}{B}. Since uu covers ∣B∣\lvert B \rvert times as many periods, there are usually about ∣B∣\lvert B \rvert times as many solutions.

In plain words Solve for the inner angle first, over the range of values the inner angle actually takes, and only then go back to xx.

Example Solve sin⁡2x=−32\sin 2x = -\frac{\sqrt{3}}{2} on [0,2π][0, 2\pi]. With u=2xu = 2x the inner angle runs over [0,4π][0, 4\pi], two full turns. In that interval

u=4π3, 5π3, 10π3, 11π3.u = \frac{4\pi}{3}, \ \frac{5\pi}{3}, \ \frac{10\pi}{3}, \ \frac{11\pi}{3}.

Dividing by 22 gives four solutions, x=2π3x = \frac{2\pi}{3}, 5π6\frac{5\pi}{6}, 5π3\frac{5\pi}{3} and 11π6\frac{11\pi}{6}. Stopping at uu in [0,2π][0, 2\pi] would have missed the last two.

Checking solutions

Definition A step that is not reversible can change the solution set. A substitution such as c=cos⁡xc = \cos x that turns the equation into a polynomial equation in cc can produce values with ∣c∣>1\lvert c \rvert > 1, which no real angle gives, and these must be discarded. Squaring both sides can add extraneous solutions, and dividing both sides by an expression such as sin⁡x\sin x loses the solutions where that expression is 00. So check every candidate in the original equation, and check that it lies in the stated interval, endpoints included.

In plain words Algebra can hand you numbers that are not real answers, and it can quietly drop real ones. A final check against the original equation and the interval catches both.

Example Solve 2cos⁡2x−5cos⁡x+2=02\cos^2 x - 5\cos x + 2 = 0 on [0,2π)[0, 2\pi). With c=cos⁡xc = \cos x it factors as (2c−1)(c−2)=0(2c - 1)(c - 2) = 0, so c=12c = \frac{1}{2} or c=2c = 2. The value 22 is impossible for a cosine and is discarded. From cos⁡x=12\cos x = \frac{1}{2} we get x=π3x = \frac{\pi}{3} and x=5π3x = \frac{5\pi}{3}. As a check at x=π3x = \frac{\pi}{3},

2(12)2−5⋅12+2=12−52+2=0.\begin{aligned} 2\left(\frac{1}{2}\right)^2 - 5 \cdot \frac{1}{2} + 2 &= \frac{1}{2} - \frac{5}{2} + 2 \\[0.4em] &= 0. \end{aligned}

Check yourself

  1. Find all solutions of cos⁡3x=0\cos 3x = 0 in [0,π][0, \pi].
Show the answer

With u=3xu = 3x the inner angle runs over [0,3π][0, 3\pi]. Cosine is zero at the odd multiples of π2\frac{\pi}{2}, and those in [0,3π][0, 3\pi] are π2\frac{\pi}{2}, 3π2\frac{3\pi}{2} and 5π2\frac{5\pi}{2}. Dividing by 33 gives x=π6x = \frac{\pi}{6}, π2\frac{\pi}{2} and 5π6\frac{5\pi}{6}.

  1. Solve 2sin⁡(π3−x)=12\sin\left(\frac{\pi}{3} - x\right) = 1 on [0,2π][0, 2\pi].
Show the answer

Put u=π3−xu = \frac{\pi}{3} - x. Here B=−1B = -1, so as xx runs over [0,2π][0, 2\pi] the inner angle runs over [−5π3,π3]\left[-\frac{5\pi}{3}, \frac{\pi}{3}\right]. The equation is sin⁡u=12\sin u = \frac{1}{2}, with general solution u=π6+2kπu = \frac{\pi}{6} + 2k\pi or u=5π6+2kπu = \frac{5\pi}{6} + 2k\pi. In the interval only u=π6u = \frac{\pi}{6} and u=−7π6u = -\frac{7\pi}{6} remain. Then x=π3−ux = \frac{\pi}{3} - u gives

x=π6andx=3π2.x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{3\pi}{2}.

As a check at x=3π2x = \frac{3\pi}{2}, the left side is 2sin⁡(−7π6)=2⋅12=12\sin\left(-\frac{7\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1. Both endpoints x=0x = 0 and x=2πx = 2\pi give 2sin⁡π3=32\sin \frac{\pi}{3} = \sqrt{3}, so neither is a solution.

  1. Solve 2sin⁡2x+3sin⁡x−2=02\sin^2 x + 3\sin x - 2 = 0 on [0,2π)[0, 2\pi).
Show the answer

With s=sin⁡xs = \sin x the equation factors as (2s−1)(s+2)=0(2s - 1)(s + 2) = 0, so s=12s = \frac{1}{2} or s=−2s = -2. A sine is never −2-2, so only sin⁡x=12\sin x = \frac{1}{2} is left, which gives x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6}. Both satisfy the original equation, since 2⋅14+32−2=02 \cdot \frac{1}{4} + \frac{3}{2} - 2 = 0.

How this lesson connects