Trigonometric Formulae, Equations and Inequalities

Lessons

Senior 5 · Mathematics

Trigonometric Formulae, Equations and Inequalities

5 lessons

Lesson 3 of 5

Solving trigonometric equations

25 min

How many angles have a sine of $\frac{1}{2}$ ?

A calculator gives one answer: $\sin^{-1} 0.5 = 30^\circ$ . But $\sin 150^\circ = \frac{1}{2}$ too, and so do $390^\circ$ , $510^\circ$ , $-210^\circ$ and infinitely many more, because the sine graph repeats every $360^\circ$ . Solving a trigonometric equation means finding all of them: first as one formula with a whole number $k$ , then the ones that lie in the interval the question asks for. The unit circle shows where the second solution comes from.

Figure 1. A horizontal line meets the unit circle at two angles with the same sine; a vertical line meets it at two angles with the same cosine. Every whole turn

2k\pi

adds more solutions.

Definition

Principal value: the one solution the calculator gives, taken from a fixed interval: $\sin^{-1} k \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , $\cos^{-1} k \in [0, \pi]$ and $\tan^{-1} b \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

General solution: a formula, containing an arbitrary integer $k \in \mathbb{Z}$ , that gives every solution of the equation and nothing else.

Example: for $\tan\theta = 1$ , the principal value is $\frac{\pi}{4}$ and the general solution is $\theta = \frac{\pi}{4} + k\pi$ , $k \in \mathbb{Z}$ .

Important

The three basic equations ( $k \in \mathbb{Z}$ ; take $\alpha$ as the principal value)

\sin\theta = \sin\alpha \iff \theta = \alpha + 2k\pi \ \text{ or } \ \theta = \pi - \alpha + 2k\pi

\cos\theta = \cos\alpha \iff \theta = \alpha + 2k\pi \ \text{ or } \ \theta = -\alpha + 2k\pi

\tan\theta = \tan\alpha \iff \theta = \alpha + k\pi

$\sin\theta = k$ and $\cos\theta = k$ have solutions only when $-1 \le k \le 1$ ; $\tan\theta = b$ has solutions for every real $b$ . The tangent needs only one family because its period is $π \pi$ .

Equations of the forms $\sin(x + a) = k$ , $\cos(x + a) = k$ and $\tan(x + a) = b$

Treat the whole bracket as one angle $\theta = x + a$ . Solve for $\theta$ with the general solution, then subtract $a$ . Only at the end pick the values of $k$ that put $x$ in the required interval.

Worked example 1: a shifted sine

Solve $\sin\left(x - \frac{\pi}{3}\right) = \frac{1}{2}$ for $0 \le x \lt 2\pi$ .

Solution The principal value is $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ , so

x - \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x - \frac{\pi}{3} = \pi - \frac{\pi}{6} + 2k\pi,

x = \frac{\pi}{2} + 2k\pi \quad \text{or} \quad x = \frac{7\pi}{6} + 2k\pi.

With $k = 0$ both lie in $[0, 2\pi)$ ; any other $k$ takes them outside. Answer: $x = \frac{\pi}{2}$ or $x = \frac{7\pi}{6}$ . Check: $\sin\left(\frac{7\pi}{6} - \frac{\pi}{3}\right) = \sin\frac{5\pi}{6} = \frac{1}{2}$ . ✓

Worked example 2: a tangent in degrees

Solve $\tan(x + 20^\circ) = 2.5$ for $0^\circ \le x \lt 360^\circ$ , to one decimal place.

Solution $\tan^{-1} 2.5 = 68.20^\circ$ , and the tangent repeats every $180^\circ$ :

x + 20^\circ = 68.20^\circ + 180^\circ k \quad \Rightarrow \quad x = 48.20^\circ + 180^\circ k.

$k = 0$ gives $48.2^\circ$ and $k = 1$ gives $228.2^\circ$ . ( $k = - 1$ gives a negative angle and $k = 2$ gives more than $360^\circ$ .) Answer: $x \approx 48.2^\circ$ or $228.2^\circ$ .

Equations of the form $\sin nx = k$

Solve for the angle $n x$ first, then divide everything, including the $2k\pi$ , by $n$ . The period becomes $\frac{2\pi}{n}$ , so the solutions come closer together. If $x$ runs over an interval of length $2\pi$ , then $n x$ runs over an interval of length $2n\pi$ : expect $n$ times as many solutions.

Worked example 3: six solutions

Solve $\sin 3x = \frac{1}{2}$ for $0 \le x \lt 2\pi$ .

Solution

3x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 3x = \frac{5\pi}{6} + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{18} + \frac{2k\pi}{3} \quad \text{or} \quad x = \frac{5\pi}{18} + \frac{2k\pi}{3}.

Since $\frac{2\pi}{3} = \frac{12\pi}{18}$ , the values $k = 0, 1, 2$ give

x = \frac{\pi}{18},\ \frac{13\pi}{18},\ \frac{25\pi}{18} \quad \text{and} \quad x = \frac{5\pi}{18},\ \frac{17\pi}{18},\ \frac{29\pi}{18}.

$k = 3$ gives $\frac{37\pi}{18} \gt 2\pi$ , too big. Figure 2 shows all six.

Figure 2.

\sin 3x

runs through three full waves on

[0, 2\pi)

, so the line

y = \frac{1}{2}

meets it

3 \times 2 = 6

times.

Warning

Divide the $2k\pi$ as well. A common error in Worked example 3 is to write $x = \frac{\pi}{18} + 2k\pi$ , which finds only 2 of the 6 solutions. Another: never divide both sides of an equation by $\sin x$ or $\cos x$ . In $\sin 2x = \sin x$ , dividing $2\sin x\cos x = \sin x$ by $\sin x$ throws away every solution of $\sin x = 0$ . Factorise instead.

Equations that reduce to the basic forms

Many equations become basic ones after an identity: use $\sin^2 x + \cos^2 x = 1$ to make a quadratic in one ratio, a double-angle formula to factorise, or a sum-to-product formula (Lesson 2) to turn a sum into a product that can be set to zero.

Worked example 4: a quadratic in $\sin x$

Solve $2\cos^2 x + 3\sin x - 3 = 0$ for $0 \le x \lt 2\pi$ .

Solution Replace $\cos^2 x$ by $1 - \sin^2 x$ : $2 - 2\sin^2 x + 3\sin x - 3 = 0$ , so $2\sin^2 x - 3\sin x + 1 = 0$ , which factorises as

(2\sin x - 1)(\sin x - 1) = 0.

$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ ; $\sin x = 1$ gives $x = \frac{\pi}{2}$ . Answer: $x \in \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\right\}$ .

Worked example 5: a sum set to zero

Solve $\sin 3x + \sin x = 0$ for $0 \le x \lt 2\pi$ .

Solution Sum into product: $2\sin 2x\cos x = 0$ , so $\sin 2x = 0$ or $\cos x = 0$ .

$\sin 2x = 0$ : $2x = k\pi$ , $x = \frac{k\pi}{2}$ , giving $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ .
$\cos x = 0$ : $x = \frac{\pi}{2}$ or $\frac{3\pi}{2}$ , already in the list.

Answer: $x \in \left\{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\right\}$ . Check another way: $\sin 3x = -\sin x = \sin(-x)$ gives $3x = -x + 2k\pi$ or $3x = \pi + x + 2k\pi$ , that is $x = \frac{k\pi}{2}$ or $x = \frac{\pi}{2} + k\pi$ : the same set. ✓

Tip

A three-step routine: (1) reduce to $\sin(\text{angle}) = \sin\alpha$ , $\cos(\text{angle}) = \cos\alpha$ or $\tan(\text{angle}) = \tan\alpha$ ; (2) write the general solution for the angle, then solve for $x$ ; (3) list the values of $k$ that land in the interval, and check one answer in the original equation.

The Organic Chemistry Tutor: Solving Trigonometric Equations By Finding All Solutions (about 25 min). Watch the first part for writing all solutions with a whole number k, and how a multiple angle such as 2x doubles the number of answers in one turn.

✦

Practice

Give exact answers unless told otherwise. The last two are harder.

Write the general solution of (a) $\sin\theta = \frac{\sqrt{3}}{2}$ (b) $\cos\theta = -1$ (c) $\tan\theta = -1$ .
Solve $\cos\left(x + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ for $0 \le x \lt 2\pi$ .
Solve $\tan(x - 15^\circ) = 1$ for $0^\circ \le x \lt 360^\circ$ .
Solve $\sin 2x = -\frac{\sqrt{3}}{2}$ for $0 \le x \lt 2\pi$ .
Solve $\cos 3x = 0$ for $0 \le x \le \pi$ .
Solve $2\sin^2 x - \sin x - 1 = 0$ for $0 \le x \lt 2\pi$ .
Solve $\sin 2x = \cos x$ for $0 \le x \lt 2\pi$ .
Solve $\cos 5x + \cos x = 0$ for $0 \le x \le \pi$ .

Show answersHide answers

(a) $\theta = \frac{\pi}{3} + 2k\pi$ or $\frac{2\pi}{3} + 2k\pi$ (b) $\theta = \pi + 2k\pi$ (c) $\theta = -\frac{\pi}{4} + k\pi$ .
$x + \frac{\pi}{6} = \pm\frac{\pi}{6} + 2k\pi$ , so $x = 2k\pi$ or $x = -\frac{\pi}{3} + 2k\pi$ : $x = 0$ or $\frac{5\pi}{3}$ .
$x - 15^\circ = 45^\circ + 180^\circ k$ : $x = 60^\circ$ or $240^\circ$ .
$2x = -\frac{\pi}{3} + 2k\pi$ or $\frac{4\pi}{3} + 2k\pi$ , so $x = -\frac{\pi}{6} + k\pi$ or $\frac{2\pi}{3} + k\pi$ : $x \in \left\{\frac{2\pi}{3}, \frac{5\pi}{6}, \frac{5\pi}{3}, \frac{11\pi}{6}\right\}$ .
$3x = \frac{\pi}{2} + k\pi$ , $x = \frac{\pi}{6} + \frac{k\pi}{3}$ : $\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$ .
$(2\sin x + 1)(\sin x - 1) = 0$ : $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$ .
$\cos x\,(2\sin x - 1) = 0$ : $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ .
$2\cos 3x\cos 2x = 0$ . $\cos 3x = 0$ : $\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$ ; $\cos 2x = 0$ : $\frac{\pi}{4}, \frac{3\pi}{4}$ . All five: $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{6}$ .

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the quiz questions.
Video: Solving Trigonometric Equations By Finding All Solutions, The Organic Chemistry Tutor, embedded from YouTube.

1How many angles have a sine of 12 \frac{1}{2} 21​?

2Equations of the forms sin⁡(x+a)=k \sin(x + a) = k sin(x+a)=k, cos⁡(x+a)=k \cos(x + a) = k cos(x+a)=k and tan⁡(x+a)=b \tan(x + a) = b tan(x+a)=b

3Equations of the form sin⁡nx=k \sin nx = k sinnx=k

4Equations that reduce to the basic forms

How many angles have a sine of $\frac{1}{2}$ ?

Equations of the forms $\sin(x + a) = k$ , $\cos(x + a) = k$ and $\tan(x + a) = b$

Equations of the form $\sin nx = k$

Equations that reduce to the basic forms