Equations with a multiple angle

Warm-up

Three quick questions from earlier lessons.

Discover

$\sin 2x$ repeats twice as fast as $\sin x$ . On $[0, 2\pi)$ , how many solutions does $\sin 2x = \dfrac{1}{2}$ have?

Four
Two

Key idea

Solve for the whole angle first, on the stretched interval.

For

\sin(nx) = c

with

0 \le x < 2\pi

: let

\theta = nx

, which runs over

[0, 2n\pi)

. Find every

\theta

there (add

2\pi

again and again), then divide each by

n

.

Worked example

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

\cos\theta = 0

:

\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \dfrac{7\pi}{2}

θ = 2x runs over [0, 4π).

x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}

Divide each by 2.

Your turn, guided

Your turn: $\sin 2x = 1$ .

\theta = \dfrac{\pi}{2}

or

\dfrac{5\pi}{2}

θ = 2x in [0, 4π): sin θ = 1 at π/2 and π/2 + 2π.

x = \dfrac{\pi}{4}

or

x = \dfrac{5\pi}{4}

Divide 5π/2 by 2.

Practice ★

How many solutions does $\sin(5x) = \dfrac{1}{2}$ have for $0 \le x < 2\pi$ ?
How many solutions does $\sin(4x) = \dfrac{1}{2}$ have for $0 \le x < 2\pi$ ?
How many solutions does $\sin(2x) = \dfrac{1}{2}$ have for $0 \le x < 2\pi$ ?

Watch out!

Practice ★★

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

Solve $\cos 3x = -1$ for $0 \le x < 2\pi$ .

Exit ticket

Three questions to check you have got it.

How many solutions does $\sin(5x) = \dfrac{1}{2}$ have for $0 \le x < 2\pi$ ?
How many solutions does $\sin(4x) = \dfrac{1}{2}$ have for $0 \le x < 2\pi$ ?
How many solutions does $\sin(2x) = \dfrac{1}{2}$ have for $0 \le x < 2\pi$ ?

Summary card

Multiple angles

Let $\theta = nx$ : it runs over $[0, 2n\pi)$ .
List every $\theta$ there.
Divide each by $n$ .

Watch out: $\sin 2x$ has twice as many solutions as $\sin x$ .

About this lesson