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 279: 38


The solution set is $$\{180^\circ+720^\circ n,n\in Z\}$$

Work Step by Step

$$\sin\frac{\theta}{2}=1$$ 1) First, we solve the equation over the interval $[0^\circ,360^\circ)$ - For $\sin\frac{\theta}{2}=1$, over the interval $[0^\circ, 360^\circ)$, there is one value of $\theta$ where $\sin\frac{\theta}{2}=1$, which is $90^\circ$. Therefore, $$\frac{\theta}{2}=\{90^\circ\}$$ (Be careful that the angle we are solving the equation for is $\frac{\theta}{2}$, not $\theta$) 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 $\frac{\theta}{2}$. $$\frac{\theta}{2}=\{90^\circ+360^\circ n,n\in Z\}$$ Finally, we find the solutions for $\theta$, which is also the solution set: $$\theta=\{180^\circ+720^\circ n,n\in Z\}$$
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