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Precalculus · Lesson 6.3

Solving Trigonometric Equations

How it works

Use a reference angle and all compatible quadrants, then restrict to the stated interval. Factoring is safer than dividing by a trigonometric expression that might be zero. A general solution includes periodic repeats.

sin⁡x=sin⁡α ⇒ x=α+2kπ or x=π−α+2kπ\sin x=\sin\alpha\ \Rightarrow\ x=\alpha+2k\pi\ \text{or}\ x=\pi-\alpha+2k\pi

Worked examples

Example 1

Solve sin⁡x=1/2\displaystyle \sin x=1/2 on [0,2π)\displaystyle [0,2\pi).

The solutions are π/6\displaystyle \pi/6 and 5π/6\displaystyle 5\pi/6.

Example 2

Solve 2cos⁡2x−1=0\displaystyle 2\cos^2x-1=0 on [0,2π)\displaystyle [0,2\pi).

Cosine is ±2/2\displaystyle \pm\sqrt{2}/2, giving π/4,3π/4,5π/4,7π/4\displaystyle \pi/4,3\pi/4,5\pi/4,7\pi/4.

Problems

Problem 6.3.1

Solve sin⁡x=3/2\displaystyle \sin x=\sqrt{3}/2 on [0,2π)\displaystyle [0,2\pi).

Unit circle with the terminal radius at 120 degrees in quadrant II.
Cosine is the horizontal coordinate and sine the vertical coordinate.
Show worked solution

The reference angle is π/3\displaystyle \pi/3. Sine is positive in quadrants I and II: x=π/3,2π/3\displaystyle x=\pi/3,2\pi/3.

Unit circle with the terminal radius at 120 degrees in quadrant II. The solution adds labeled relationships.
The terminal point is (-1/2, sqrt(3)/2); tangent is -sqrt(3).

Problem 6.3.2

Solve sin⁡xcos⁡x=0\displaystyle \sin x\cos x=0 on [0,2π)\displaystyle [0,2\pi).

Show worked solution

Use the zero-product property. The solutions are 0,π/2,π,3π/2\displaystyle 0,\pi/2,\pi,3\pi/2.

Problem 6.3.3

Give all real solutions to tan⁡x=1\displaystyle \tan x=1.

Show worked solution

Tangent has period π\displaystyle \pi, so x=π/4+kπ\displaystyle x=\pi/4+k\pi for any integer k.

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