Solving Linear Trigonometric Equations, precalculus lesson

The height of a seat on a turning wheel rises and falls like a sine curve. Asking when the seat reaches a certain height leads to an equation such as $2\sin x + 1 = 0$ . An inverse function gives one answer, but the wheel keeps turning, so the seat passes that height twice on every turn. Solving a trigonometric equation means finding all of those moments, not just the first one.

Key terms

Isolating the trigonometric function

Definition A linear trigonometric equation is one that can be rearranged into the form

a \, T(x) + b = 0, \qquad a \neq 0,

where $T$ is one of sine, cosine, tangent, cotangent, secant or cosecant, applied to the unknown $x$ . Isolating the function gives $T(x) = c$ with $c = -\frac{b}{a}$ . Solutions exist only when $c$ lies in the range of $T$ , which is $[-1, 1]$ for sine and cosine, all real numbers for tangent and cotangent, and $(-\infty, -1] \cup [1, \infty)$ for secant and cosecant.

In plain words Treat $\sin x$ , or whichever function appears, as a single unknown and solve for it exactly as in a linear equation. Then ask whether the function can actually take that value.

Example In $5\sin x - 2 = 3\sin x - 3$ , collect the sines on the left to get $2\sin x = -1$ , so $\sin x = -\frac{1}{2}$ . That value is in $[-1, 1]$ , so solutions exist. By contrast, the equation $3\cos x - 4 = 0$ leads to $\cos x = \frac{4}{3}$ , which is larger than $1$ , so it has no solution at all.

Principal values

Definition For an equation $\sin x = c$ , $\cos x = c$ or $\tan x = c$ , the principal value is the single solution that the matching inverse function returns. In radians,

\begin{gathered} \arcsin c \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], \\[0.8em] \arccos c \in [0, \pi], \\[0.4em] \arctan c \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \end{gathered}

where $\arcsin c$ and $\arccos c$ need $-1 \le c \le 1$ and $\arctan c$ accepts every real $c$ . A calculator in radian mode gives decimal approximations of these same values.

In plain words The inverse function hands you one particular answer, always taken from the same fixed interval. It is the starting point of the solution, not the whole solution.

Example Here are three principal values.

Equation	Principal value
$\sin x = -\frac{1}{2}$	$\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$
$\tan x = 3$	$\arctan 3 \approx 1.2490$
$\cos x = -0.4$	$\arccos(-0.4) \approx 1.9823$

General periodic solutions

Definition Let $k$ stand for any integer. For $-1 \le c \le 1$ ,

\begin{aligned} &\sin x = c \iff x = \arcsin c + 2k\pi \\[0.4em] &\quad \text{or } x = \pi - \arcsin c + 2k\pi, \end{aligned}

\begin{aligned} &\cos x = c \iff x = \arccos c + 2k\pi \\[0.4em] &\quad \text{or } x = -\arccos c + 2k\pi. \end{aligned}

For every real $c$ ,

\tan x = c \iff x = \arctan c + k\pi.

When $c = \pm 1$ , the two families for sine describe the same numbers, and so do the two families for cosine.

In plain words Sine and cosine take each value strictly between $-1$ and $1$ twice per turn, at two angles that mirror each other on the unit circle, across the vertical axis for sine and across the horizontal axis for cosine. Then everything repeats every $2\pi$ . Tangent takes each value once per half turn and repeats every $\pi$ .

Example For $\sin x = -\frac{1}{2}$ the principal value is $-\frac{\pi}{6}$ , and the mirror angle is $\pi - \left( -\frac{\pi}{6} \right) = \frac{7\pi}{6}$ . So the general solution is

\begin{aligned} x &= -\frac{\pi}{6} + 2k\pi \quad \text{or} \\[0.8em] x &= \frac{7\pi}{6} + 2k\pi, \qquad k \in \mathbb{Z}. \end{aligned}

All solutions in an interval

Definition To find every solution of a trigonometric equation in an interval $I$ , write its general solution and, in each family, try the integers $k$ one after another, keeping exactly the values that lie in $I$ . Values in one family are a full period apart, so a bounded interval contains only finitely many of them, and once a family has left $I$ in both directions no further $k$ can bring it back.

In plain words Step through $k = 0, 1, 2, \dots$ and $k = -1, -2, \dots$ in each family until the values leave the interval. Every value inside the interval counts, and none may be skipped.

Example Solve $2\sin x + 1 = 0$ on $[0, 4\pi]$ . The general solution from the previous section gives the table below. Writing $4\pi = \frac{24\pi}{6}$ makes it easy to see which values fit.

$k$	$-\frac{\pi}{6} + 2k\pi$	$\frac{7\pi}{6} + 2k\pi$
$0$	$-\frac{\pi}{6}$ , not in the interval	$\frac{7\pi}{6}$
$1$	$\frac{11\pi}{6}$	$\frac{19\pi}{6}$
$2$	$\frac{23\pi}{6}$	$\frac{31\pi}{6}$ , not in the interval

Every negative $k$ gives values below $0$ in both families. So the solutions are $\frac{7\pi}{6}$ , $\frac{11\pi}{6}$ , $\frac{19\pi}{6}$ and $\frac{23\pi}{6}$ , two for each turn.

Secant, cosecant and cotangent equations

Definition For every real $c \neq 0$ ,

\begin{gathered} \sec x = c \iff \cos x = \frac{1}{c}, \\[0.8em] \csc x = c \iff \sin x = \frac{1}{c}, \\[0.8em] \cot x = c \iff \tan x = \frac{1}{c}. \end{gathered}

The equation $\cot x = 0$ holds exactly when $\cos x = 0$ , that is, $x = \frac{\pi}{2} + k\pi$ with $k$ an integer. The equations $\sec x = 0$ and $\csc x = 0$ have no solution.

In plain words Flip the equation over to the function you already know how to solve. The angles you find automatically avoid the places where the original function is undefined.

Example Solve $\sec x + 2 = 0$ on $[0, 2\pi)$ . Isolating gives $\sec x = -2$ , so $\cos x = -\frac{1}{2}$ . The principal value is $\arccos\left( -\frac{1}{2} \right) = \frac{2\pi}{3}$ , and its mirror angle in the interval is $2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ . Both have $\cos x \neq 0$ , so the solutions are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ .

Excluding undefined inputs

Definition A solution of an equation must lie in the domain of every expression in it. Tangent and secant are undefined at $x = \frac{\pi}{2} + k\pi$ , and cotangent and cosecant at $x = k\pi$ , for every integer $k$ . Simplifying with an identity such as $\tan x = \frac{\sin x}{\cos x}$ , or multiplying both sides by an expression that can be zero, may produce an equation that holds at inputs where the original is undefined. Those inputs must be removed.

In plain words Simplifying can quietly remove a denominator. Any answer that makes an original denominator zero was never a solution.

Example Solve $\tan x \cos x = 1$ . Wherever $\cos x \neq 0$ the left side simplifies to $\sin x$ , so the equation becomes $\sin x = 1$ , with candidates $x = \frac{\pi}{2} + 2k\pi$ . At every one of these inputs $\cos x = 0$ , so $\tan x$ is undefined there. The original equation has no solution.

Check yourself

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

Show the answer

Isolating gives $\cos x = \frac{\sqrt{3}}{2}$ . The principal value is $\frac{\pi}{6}$ , and the mirror angle in the interval is $2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ . The solutions are $\frac{\pi}{6}$ and $\frac{11\pi}{6}$ .

Find the general solution of $\sqrt{3} \tan x + 1 = 0$ .

Show the answer

Isolating gives $\tan x = -\frac{1}{\sqrt{3}}$ , with principal value $\arctan\left( -\frac{1}{\sqrt{3}} \right) = -\frac{\pi}{6}$ . Tangent repeats every $\pi$ , so

x = -\frac{\pi}{6} + k\pi, \qquad k \in \mathbb{Z}.

Solve $4\sin x + 1 = 0$ on $[0, 2\pi)$ , rounding to four decimal places.

Show the answer

Isolating gives $\sin x = -\frac{1}{4}$ , with principal value $\arcsin\left( -\frac{1}{4} \right) \approx -0.2527$ . That value is outside the interval, so add $2\pi$ to get about $6.0305$ . The mirror angle is $\pi + 0.2527 \approx 3.3943$ . The solutions are about $3.3943$ and $6.0305$ .

Solving Linear Trigonometric Equations

Key terms

Isolating the trigonometric function

Principal values

General periodic solutions

All solutions in an interval

Secant, cosecant and cotangent equations

Excluding undefined inputs

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus