Solve trigonometric equations by factoring

About this lesson

Precalculus › Trigonometric identities and equations

To save a PDF, choose “Save as PDF” in the print dialog.

Solve trigonometric equations by factoring

Warm-up

Three quick questions from earlier lessons.

Discover

To solve sin⁡xcos⁡x=sin⁡x\sin x \cos x = \sin x, Leo divided both sides by sin⁡x\sin x. Is that safe?

  1. No: it loses the solutions where sin⁡x=0\sin x = 0
  2. Yes, it is always allowed

Key idea

Treat sin⁡x\sin x or cos⁡x\cos x like an unknown, then factor.

Bring every term to one side and factor (never divide by sin⁡x\sin x or cos⁡x\cos x). A product is 0 when a factor is 0. For a quadratic in sin⁡x\sin x, let u=sin⁡xu = \sin x, solve for uu, then solve each sin⁡x=u\sin x = u (no solution if ∣u∣>1|u| > 1).

Worked example

Solve 2sin⁡2x−sin⁡x=02\sin^2 x - \sin x = 0 for 0≤x<2π0 \le x < 2\pi.

sin⁡x (2sin⁡x−1)=0\sin x\,(2\sin x - 1) = 0
Common factor sin x.
sin⁡x=0\sin x = 0: x=0x = 0 or x=πx = \pi
First factor.
x=π6x = \dfrac{\pi}{6} or x=5π6x = \dfrac{5\pi}{6}
Second factor: sin x = 1/2.

Your turn, guided

Your turn: cos⁡2x−1=0\cos^2 x - 1 = 0.

cos⁡x=1\cos x = 1 or cos⁡x=−1\cos x = -1
cos²x = 1.
x=0x = 0 or x=πx = \pi
cos x = −1: type the angle.

Practice ★

  1. Solve sin⁡x(sin⁡x−32)=0\sin x\left(\sin x - \frac{\sqrt{3}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡x(sin⁡x−22)=0\sin x\left(\sin x - \frac{\sqrt{2}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  3. Solve sin⁡x(sin⁡x−12)=0\sin x\left(\sin x - \frac{1}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.

Watch out!

Practice ★★

  1. Solve 3cos⁡(x)+cos⁡(2x)+2=03 \cos{\left(x \right)} + \cos{\left(2 x \right)} + 2 = 0 for 0≤x<2π0 \le x < 2\pi.
  2. Solve −3cos⁡(x)+cos⁡(2x)+2=0- 3 \cos{\left(x \right)} + \cos{\left(2 x \right)} + 2 = 0 for 0≤x<2π0 \le x < 2\pi.
  1. Put the steps for solving 2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 0 in order.
    • Let u=sin⁡xu = \sin x: 2u2+u−1=02u^2 + u - 1 = 0
    • Factor: (2u−1)(u+1)=0(2u - 1)(u + 1) = 0
    • sin⁡x=12\sin x = \dfrac{1}{2} or sin⁡x=−1\sin x = -1
    • List every xx in [0,2π)[0, 2\pi): π6,5π6,3π2\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}

Exit ticket

Three questions to check you have got it.

  1. Solve sin⁡x(sin⁡x−32)=0\sin x\left(\sin x - \frac{\sqrt{3}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡x(sin⁡x−22)=0\sin x\left(\sin x - \frac{\sqrt{2}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  3. Solve sin⁡x(sin⁡x−12)=0\sin x\left(\sin x - \frac{1}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.

Summary card

Solving by factoring

  1. One side 0, then factor (or set u=sin⁡xu = \sin x).
  2. Each factor = 0.
  3. Solve each sin⁡x=u\sin x = u in [0,2π)[0, 2\pi).

Watch out: never divide by sin⁡x\sin x or cos⁡x\cos x.

End of the lesson

Next lesson: Equations with a multiple angle Back to the chapter

Answers: Solve trigonometric equations by factoring

A

  1. 0, π, π3, 2π30,\ \pi,\ \frac{\pi}{3},\ \frac{2 \pi}{3}
  2. 0, π, π4, 3π40,\ \pi,\ \frac{\pi}{4},\ \frac{3 \pi}{4}
  3. 0, π, π6, 5π60,\ \pi,\ \frac{\pi}{6},\ \frac{5 \pi}{6}

B

  1. π, 2π3, 4π3\pi,\ \frac{2 \pi}{3},\ \frac{4 \pi}{3}
  2. 0, π3, 5π30,\ \frac{\pi}{3},\ \frac{5 \pi}{3}

C

  1. 1. Let u=sin⁡xu = \sin x: 2u2+u−1=02u^2 + u - 1 = 0 2. Factor: (2u−1)(u+1)=0(2u - 1)(u + 1) = 0 3. sin⁡x=12\sin x = \dfrac{1}{2} or sin⁡x=−1\sin x = -1 4. List every xx in [0,2π)[0, 2\pi): π6,5π6,3π2\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}

D

  1. 0, π, π3, 2π30,\ \pi,\ \frac{\pi}{3},\ \frac{2 \pi}{3}
  2. 0, π, π4, 3π40,\ \pi,\ \frac{\pi}{4},\ \frac{3 \pi}{4}
  3. 0, π, π6, 5π60,\ \pi,\ \frac{\pi}{6},\ \frac{5 \pi}{6}

CC BY-SA 4.0