Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 280: 41


The solution set is $$\{30^\circ+360^\circ n,150^\circ+360^\circ n,270^\circ+360^\circ n,n\in Z\}$$

Work Step by Step

$$2\sin\theta=2\cos2\theta$$ Here we have two different trigonometric functions, sine and cosine. It would be better if we could rewrite the equation in only one trigonometric function. Luckily, there is a way to do so. - Recall the identity: $\cos2\theta=1-2\sin^2\theta$ which means $$2\sin\theta=2(1-2\sin^2\theta)$$ $$2\sin\theta=2-4\sin^2\theta$$ $$4\sin^2\theta+2\sin\theta-2=0$$ $$2\sin^2\theta+\sin\theta-1=0$$ $$(2\sin^2\theta+2\sin\theta)+(-\sin\theta-1)=0$$ $$2\sin\theta(\sin\theta+1)-(\sin\theta+1)=0$$ $$(\sin\theta+1)(2\sin\theta-1)=0$$ $$\sin\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=\frac{1}{2}$$ 1) First, we solve the equation over the interval $[0^\circ,360^\circ)$ - First, $\sin\theta=-1$, over the interval $[0^\circ, 360^\circ)$, there is one value of $\theta$ where $\sin\theta=-1$, which are $\theta=\{270^\circ\}$ - For $\sin\theta=\frac{1}{2}$, over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\sin\theta=\frac{1}{2}$, which are $\theta=\{30^\circ,150'^\circ\}$ Therefore, overall, $$\theta=\{30^\circ,150^\circ,270^\circ\}$$ 2) Solve the equation for all solutions Sine function has period $360^\circ$, so we would add $360^\circ$ to all solutions found in part 1) for $\theta$. $$\theta=\{30^\circ+360^\circ n,150^\circ+360^\circ n,270^\circ+360^\circ n,n\in Z\}$$ This is the solution set of the given equation.
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