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: 46


The solution set is $$\{1.231+2n\pi,5.0522+2n\pi, n\in Z\}$$

Work Step by Step

$$\cos x=\sin^2\frac{x}{2}$$ - Recall the identity: $\cos2x=1-2\sin^2x$ Thus, we can deduce that $$\cos x=1-2\sin^2\frac{x}{2}$$ $$2\sin^2\frac{x}{2}=1-\cos x$$ $$\sin^2\frac{x}{2}=\frac{1}{2}-\frac{\cos x}{2}$$ Apply back to the equation: $$\cos x=\frac{1}{2}-\frac{\cos x}{2}$$ $$\frac{3\cos x}{2}=\frac{1}{2}$$ $$\cos x=\frac{1}{3}$$ 1) First, we solve the equation over the interval $[0,2\pi)$ For $\cos x=\frac{1}{3}$, we have $$x=\cos^{-1}\frac{1}{3}$$ $$x\approx1.231$$ Also, over the interval $[0,2\pi)$, there is one more value of $x$ where $\cos x=\frac{1}{3}$, which is $x\approx2\pi-1.231\approx5.0522$ Therefore, $$x=\{1.231,5.0522\}$$ 2) Solve the equation for all solutions Cosine function has period $2\pi$, so we would add $2\pi$ to all solutions found in part 1) for $x$. $$x=\{1.231+2n\pi,5.0522+2n\pi, n\in Z\}$$
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