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: 45


The solution set is $$\{1.3181+2n\pi,4.9651+2n\pi, n\in Z\}$$

Work Step by Step

$$3\csc^2\frac{x}{2}=2\sec x$$ Here we encounter both functions for angle $x$ and half-angle $\frac{x}{2}$. Currently there is no available identity to deal with half-angle for secant and cosecant functions. So we need to switch them back to sine and cosine functions. - Recall the identity: $\csc\frac{x}{2}=\frac{1}{\sin\frac{x}{2}}$ and $\sec x=\frac{1}{\cos x}$ $$\frac{3}{\sin^2\frac{x}{2}}=\frac{2}{\cos x}$$ $$3\cos x=2\sin^2\frac{x}{2}$$ - Recall the identity: $\cos2x=1-2\sin^2x$ Thus, we can deduce that $$\cos x=1-2\sin^2\frac{x}{2}$$ $$2\sin^2\frac{x}{2}=1-\cos x$$ Apply back to the equation: $$3\cos x=1-\cos x$$ $$4\cos x=1$$ $$\cos x=\frac{1}{4}$$ 1) First, we solve the equation over the interval $[0,2\pi)$ For $\cos x=\frac{1}{4}$, we have $$x=\cos^{-1}\frac{1}{4}$$ $$x\approx1.3181$$ Also, over the interval, there is one more value of $x$ where $\cos x=\frac{1}{4}$, which is $x\approx2\pi-1.3181\approx4.9651$ Therefore, $$x=\{1.3181,4.9651\}$$ 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 $x$. $$x=\{1.3181+2n\pi,4.9651+2n\pi, n\in Z\}$$
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