## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 53

#### Answer

The solution set is $$\{\frac{\pi}{3}+2\pi n, \pi+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$

#### Work Step by Step

$$2\cos^2 x+\cos x-1=0$$ 1) Solve the equation over the interval $[0,2\pi)$ $$2\cos^2 x+\cos x-1=0$$ $$(2\cos^2 x+2\cos x)+(-\cos x-1)=0$$ $$2\cos x(\cos x+1)-(\cos x+1)=0$$ $$(\cos x+1)(2\cos x-1)=0$$ $$\cos x=-1\hspace{1cm}\text{or}\hspace{1cm}\cos x=\frac{1}{2}$$ - For $\cos x=-1$ Over the interval $[0,2\pi)$, there is one value of $x$ where $\cos x=-1$, which is $\pi$. - For $\cos x=\frac{1}{2}$ Over the interval $[0,2\pi)$, there is one value of $x$ where $\cos x=\frac{1}{2}$, which are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$. Therefore, overall, $$x=\{\frac{\pi}{3},\pi,\frac{5\pi}{3}\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the cosine function is $2\pi$. - We apply it to each solution found in step 1. The results are the solution set. So the solution set is $$\{\frac{\pi}{3}+2\pi n, \pi+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$

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