Solve equations with identities

Warm-up

Three quick questions from earlier lessons.

Discover

To solve $\sin 2x = \sin x$ , what is a good first step?

Replace $\sin 2x$ by $2\sin x\cos x$
Write $\sin 2x = 2\sin x$
Divide both sides by $\sin x$

Key idea

Identities turn an equation with different angles into one you can factor.

Rewrite with one angle (

\sin 2x = 2\sin x\cos x

;

\cos 2x = 2\cos^2 x - 1

or

1 - 2\sin^2 x

, whichever matches the other terms). Then bring everything to one side, factor, and solve each factor on

[0, 2\pi)

.

Worked example

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

2\sin x\cos x - \sin x = 0

Double-angle identity.

\sin x\,(2\cos x - 1) = 0

Common factor, never divide.

x = 0

or

x = \pi

sin x = 0.

x = \dfrac{\pi}{3}

or

x = \dfrac{5\pi}{3}

cos x = 1/2.

Your turn, guided

Your turn: $\cos 2x + \cos x = 0$ , using $\cos 2x = 2\cos^2 x - 1$ .

2\cos^2 x + \cos x - 1 = 0

A quadratic in cos x.

\cos x = \dfrac{1}{2}

or

\cos x = -1

(2cos x − 1)(cos x + 1) = 0: type the positive value of cos x.

Practice ★

Solve $\sin 2x = \sin{\left(x \right)}$ for $0 \le x < 2\pi$ .
Solve $\sin 2x = - \sin{\left(x \right)}$ for $0 \le x < 2\pi$ .

Watch out!

Practice ★★

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

Exit ticket

Three questions to check you have got it.

Solve $\sin 2x = \sin{\left(x \right)}$ for $0 \le x < 2\pi$ .
Solve $\sin 2x = - \sin{\left(x \right)}$ for $0 \le x < 2\pi$ .

Summary card

Equations with identities

Rewrite with one angle (double-angle identities).
One side 0, factor (or $u = \sin x$ , $u = \cos x$ ).
Solve each factor in $[0, 2\pi)$ .

Watch out: factor, never divide by $\sin x$ or $\cos x$ .

About this lesson