Solving Linear Trig Equations

A linear trig equation is one where the trig ratio appears only to the first power, like $3\cos x - 1 = 0$. You solve it in two stages: first use ordinary algebra to isolate the trig ratio, then use the unit circle to find every angle in the interval that has that ratio. It’s the same process you used in degrees in Grade 11, but now the interval is $0 \le x \le 2\pi$ and all answers are in radians.

Key ideas

Step 1: Isolate the trig ratio

Solve for $\sin x$, $\cos x$ or $\tan x$ just as you would solve for a variable:

$$a\sin x + b = c \quad\Rightarrow\quad \sin x = \frac{c - b}{a}$$

Sine and cosine are always between $-1$ and $1$. If you get $\sin x = 1.25$ or $\cos x = -2$, there is no solution. (Tangent can be any real number.)

Step 2: Find the related (reference) angle

The related angle $\theta_r$ is the acute angle between the terminal arm and the $x$-axis. Find it from the positive value of the ratio:

• If it’s a special value ($\tfrac{1}{2}$, $\tfrac{\sqrt{2}}{2}$, $\tfrac{\sqrt{3}}{2}$, $1$, $\sqrt{3}$, $\tfrac{1}{\sqrt{3}}$), use the special angles for an exact answer.
• Otherwise, use the inverse function on your calculator in radian mode: for example, $\theta_r = \sin^{-1}(0.4) \approx 0.4115$.

Step 3: Use CAST to place the angles

The sign of the ratio tells you which quadrants the answers are in. Then:

Quadrant	1	2	3	4
Angle	$\theta_r$	$\pi - \theta_r$	$\pi + \theta_r$	$2\pi - \theta_r$
Positive ratios	all	sine	tangent	cosine

Most equations have two solutions in $0 \le x \le 2\pi$. The exceptions are values on the axes: for example, $\sin x = 1$ only at $x = \dfrac{\pi}{2}$, and $\sin x = 0$ at $x = 0$, $\pi$ and $2\pi$ (both endpoints count, since the interval includes them).

Equations with kx

For an equation like $\cos 2x = \dfrac{1}{2}$ on $0 \le x \le 2\pi$, the angle is $2x$, and it runs over $0 \le 2x \le 4\pi$: that’s two full turns. So there are twice as many solutions.

1. Let $u = kx$ and find the interval for $u$: $0 \le u \le 2k\pi$.
2. Solve for $u$ in the first turn, then add $2\pi$ (and $4\pi$, …) to get the solutions in the later turns.
3. Divide every value by $k$ to get $x$.

The graph of $y = \sin kx$ or $y = \cos kx$ has period $\dfrac{2\pi}{k}$, so it completes $k$ cycles between $0$ and $2\pi$. That’s where the extra solutions come from.

Checking with a graph

Each solution of $f(x) = c$ is the $x$-coordinate of an intersection of $y = f(x)$ and the horizontal line $y = c$. Graphing both (for example, in Desmos with radians) is a quick way to check that you have the right number of solutions in roughly the right places.

Worked examples

Example 1: An exact answer

Solve $2\sin x - \sqrt{3} = 0$ for $0 \le x \le 2\pi$.

Solution. Isolate $\sin x$:

$$2\sin x = \sqrt{3} \quad\Rightarrow\quad \sin x = \frac{\sqrt{3}}{2}$$

The related angle is $\dfrac{\pi}{3}$. Sine is positive in quadrants 1 and 2:

$$x = \frac{\pi}{3} \qquad\text{or}\qquad x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

Check: $2\sin\dfrac{2\pi}{3} - \sqrt{3} = 2\left(\dfrac{\sqrt{3}}{2}\right) - \sqrt{3} = 0$. ✓

Example 2: A negative ratio

Solve $4\cos x + 3 = 1$ for $0 \le x \le 2\pi$.

Solution.

$$4\cos x = -2 \quad\Rightarrow\quad \cos x = -\frac{1}{2}$$

The related angle comes from the positive value $\dfrac{1}{2}$: $\theta_r = \dfrac{\pi}{3}$. Cosine is negative in quadrants 2 and 3:

$$x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \qquad\text{or}\qquad x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$$

Check: $4\cos\dfrac{4\pi}{3} + 3 = 4\left(-\dfrac{1}{2}\right) + 3 = 1$. ✓

Example 3: A decimal answer

Solve $5\sin x + 2 = 0$ for $0 \le x \le 2\pi$. Round to three decimal places.

Solution.

$$5\sin x = -2 \quad\Rightarrow\quad \sin x = -0.4$$

This isn’t a special value, so use the calculator (radian mode) with the positive value to get the related angle:

$$\theta_r = \sin^{-1}(0.4) \approx 0.4115$$

Sine is negative in quadrants 3 and 4:

$$\begin{aligned} x &= \pi + \theta_r \approx 3.1416 + 0.4115 \approx 3.553 \\ x &= 2\pi - \theta_r \approx 6.2832 - 0.4115 \approx 5.872 \end{aligned}$$

The line $y = -0.4$ meets $y = \sin x$ twice in $0 \le x \le 2\pi$.

Why not just press $\sin^{-1}(-0.4)$? The calculator gives about $-0.4115$, which is outside the interval and is only one of the angles. Working from the related angle finds both.

Example 4: An equation with 2x

Solve $2\cos 2x = 1$ for $0 \le x \le 2\pi$.

Solution. Isolate the cosine: $\cos 2x = \dfrac{1}{2}$.

Let $u = 2x$. Since $0 \le x \le 2\pi$, we need $0 \le u \le 4\pi$.

In the first turn, $\cos u = \dfrac{1}{2}$ at $u = \dfrac{\pi}{3}$ (quadrant 1) and $u = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$ (quadrant 4). Add $2\pi$ for the second turn:

$$u = \frac{\pi}{3},\ \frac{5\pi}{3},\ \frac{7\pi}{3},\ \frac{11\pi}{3}$$

Divide by $2$ to get $x$:

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

$y = \cos 2x$ has period $\pi$, so it reaches $\dfrac{1}{2}$ four times in $0 \le x \le 2\pi$.

Check: $2\cos\left(2\cdot\dfrac{7\pi}{6}\right) = 2\cos\dfrac{7\pi}{3} = 2\cos\dfrac{\pi}{3} = 1$. ✓

Common mistakes

Calculator in degree mode. $\sin^{-1}(0.4)$ is about $23.58$ in degrees but $0.4115$ in radians. Every answer on this page is in radians.

Taking the inverse of a negative value and stopping. $\sin^{-1}(-0.4) \approx -0.4115$ isn’t in the interval, and it’s only one solution anyway. Find the related angle from the positive value, then use CAST.

Finding only one solution. Most values of sine or cosine occur twice in one turn. Always ask “which two quadrants?”

Dividing by k too early. In $\cos 2x = \dfrac{1}{2}$, finding $x = \dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$ and stopping misses half the answers. Extend the interval for $u = 2x$ to $0 \le u \le 4\pi$ first, find all values of $u$, then divide.

Missing the endpoints. For $0 \le x \le 2\pi$, both $0$ and $2\pi$ are included. $\sin x = 0$ has three solutions: $0$, $\pi$ and $2\pi$.

Not noticing there’s no solution. If you get a value of $\sin x$ or $\cos x$ that is less than $-1$ or greater than $1$, stop: there is no solution. Your calculator will show an error, and that’s the reason.

Practice

1. (Warm-up) Solve each equation for $0 \le x \le 2\pi$.

• (a) $\sin x = -\dfrac{\sqrt{2}}{2}$
• (b) $\cos x = -1$
• (c) $\sin x = 0$

Solution

(a) Related angle $\dfrac{\pi}{4}$. Sine is negative in quadrants 3 and 4: $x = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$ or $x = 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$.

(b) The point $(-1, 0)$ on the unit circle: $x = \pi$.

(c) The points $(1, 0)$ and $(-1, 0)$: $x = 0$, $\pi$, $2\pi$.

2. (Warm-up) Solve $2\cos x - 1 = 0$ for $0 \le x \le 2\pi$.

Solution

$\cos x = \dfrac{1}{2}$. Related angle $\dfrac{\pi}{3}$; cosine is positive in quadrants 1 and 4:

$$x = \frac{\pi}{3} \qquad\text{or}\qquad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$

3. (Core) Solve $\sqrt{3}\tan x + 1 = 0$ for $0 \le x \le 2\pi$.

Solution

$\tan x = -\dfrac{1}{\sqrt{3}}$. The related angle is $\dfrac{\pi}{6}$ (since $\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$). Tangent is negative in quadrants 2 and 4:

$$x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \qquad\text{or}\qquad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$$

Check: $\tan\dfrac{11\pi}{6} = -\dfrac{1}{\sqrt{3}}$, so $\sqrt{3}\left(-\dfrac{1}{\sqrt{3}}\right) + 1 = 0$. ✓

4. (Core) Solve $3\cos x - 1 = 0$ for $0 \le x \le 2\pi$. Round to three decimal places.

Solution

$\cos x = \dfrac{1}{3}$. In radian mode, $\theta_r = \cos^{-1}\left(\dfrac{1}{3}\right) \approx 1.2310$. Cosine is positive in quadrants 1 and 4:

$$x \approx 1.231 \qquad\text{or}\qquad x \approx 2\pi - 1.2310 \approx 5.052$$

5. (Core) Explain why $4\sin x + 7 = 2$ has no solution.

Solution

Isolating gives $4\sin x = -5$, so $\sin x = -\dfrac{5}{4} = -1.25$. But $\sin x$ is always between $-1$ and $1$, so no angle has a sine of $-1.25$. (On a graph, the line $y = -1.25$ never meets $y = \sin x$.)

6. (Core) Solve $2\sin 2x + \sqrt{3} = 0$ for $0 \le x \le 2\pi$.

Solution

$\sin 2x = -\dfrac{\sqrt{3}}{2}$. Let $u = 2x$, with $0 \le u \le 4\pi$.

Related angle $\dfrac{\pi}{3}$. Sine is negative in quadrants 3 and 4, so in the first turn $u = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$ and $u = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$. Adding $2\pi$: $u = \dfrac{10\pi}{3}$ and $\dfrac{11\pi}{3}$.

Divide by $2$:

$$x = \frac{2\pi}{3},\ \frac{5\pi}{6},\ \frac{5\pi}{3},\ \frac{11\pi}{6}$$

7. (Core) A Ferris wheel at the CNE turns once every $60$ s. A rider’s height above the ground, in metres, $t$ seconds after boarding is

$$h(t) = -9\cos\left(\frac{\pi t}{30}\right) + 11$$

During the first turn ($0 \le t \le 60$), when is the rider $15.5$ m above the ground?

Solution

$$-9\cos\left(\frac{\pi t}{30}\right) + 11 = 15.5 \quad\Rightarrow\quad -9\cos\left(\frac{\pi t}{30}\right) = 4.5 \quad\Rightarrow\quad \cos\left(\frac{\pi t}{30}\right) = -\frac{1}{2}$$

Let $u = \dfrac{\pi t}{30}$. For $0 \le t \le 60$, $0 \le u \le 2\pi$. Cosine is negative in quadrants 2 and 3, with related angle $\dfrac{\pi}{3}$:

$$u = \frac{2\pi}{3} \quad\text{or}\quad u = \frac{4\pi}{3}$$

Solve $\dfrac{\pi t}{30} = u$ for $t$, so $t = \dfrac{30u}{\pi}$:

$$t = \frac{30}{\pi}\cdot\frac{2\pi}{3} = 20 \qquad\text{or}\qquad t = \frac{30}{\pi}\cdot\frac{4\pi}{3} = 40$$

The rider is $15.5$ m up at $20$ s (on the way up) and $40$ s (on the way down).

Check: $h(20) = -9\cos\dfrac{2\pi}{3} + 11 = -9\left(-\dfrac{1}{2}\right) + 11 = 15.5$. ✓

8. (Challenge) Solve $\sin 3x = \dfrac{1}{2}$ for $0 \le x \le 2\pi$. How many solutions are there, and why?

Solution

Let $u = 3x$, with $0 \le u \le 6\pi$ (three full turns).

In the first turn, $\sin u = \dfrac{1}{2}$ at $u = \dfrac{\pi}{6}$ and $u = \dfrac{5\pi}{6}$. Add $2\pi = \dfrac{12\pi}{6}$ and $4\pi = \dfrac{24\pi}{6}$:

$$u = \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6},\ \frac{17\pi}{6},\ \frac{25\pi}{6},\ \frac{29\pi}{6}$$

Divide by $3$:

$$x = \frac{\pi}{18},\ \frac{5\pi}{18},\ \frac{13\pi}{18},\ \frac{17\pi}{18},\ \frac{25\pi}{18},\ \frac{29\pi}{18}$$

There are $6$ solutions: $y = \sin 3x$ completes $3$ cycles in $0 \le x \le 2\pi$, and it reaches $\dfrac{1}{2}$ twice in each cycle.

9. (Challenge) Solve $2\cos\left(x - \dfrac{\pi}{6}\right) + \sqrt{2} = 0$ for $0 \le x \le 2\pi$.

Solution

$\cos\left(x - \dfrac{\pi}{6}\right) = -\dfrac{\sqrt{2}}{2}$. Let $u = x - \dfrac{\pi}{6}$. Since $0 \le x \le 2\pi$, the interval for $u$ is $-\dfrac{\pi}{6} \le u \le \dfrac{11\pi}{6}$.

Related angle $\dfrac{\pi}{4}$; cosine is negative in quadrants 2 and 3:

$$u = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \qquad\text{or}\qquad u = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$

Both are in the interval for $u$ (and no others are: adding or subtracting $2\pi$ takes them outside it). Add $\dfrac{\pi}{6}$ to get $x$:

$$x = \frac{3\pi}{4} + \frac{\pi}{6} = \frac{9\pi + 2\pi}{12} = \frac{11\pi}{12} \qquad\text{or}\qquad x = \frac{5\pi}{4} + \frac{\pi}{6} = \frac{15\pi + 2\pi}{12} = \frac{17\pi}{12}$$