3.6: Trigonometric Equations and Identities | AP Precalculus

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\alpha\ \Rightarrow\ x=\alpha+2k\pi\ \text{or}\ x=\pi-\alpha+2k\pi

Worked examples

Example 1

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

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

Example 2

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

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

Problems

Problem 3.6.1

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

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 $\displaystyle \pi/3$ . Sine is positive in quadrants I and II: $\displaystyle x=\pi/3,2\pi/3$ .

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 3.6.2

Solve $\displaystyle \sin x\cos x=0$ on $\displaystyle [0,2\pi)$ .

Show worked solution

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

Problem 3.6.3

Solve $\displaystyle 2\sin^2x-\sin x=0$ on $\displaystyle [0,2\pi)$ .

Show worked solution

Factor $\displaystyle \sin x(2\sin x-1)=0$ . From sine zero get $\displaystyle 0,\pi$ ; from sine one-half get $\displaystyle \pi/6,5\pi/6$ . Do not divide by sine and lose the first pair.

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 3.6.1

Problem 3.6.2

Problem 3.6.3

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