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: 51


The solution set is $$\{\frac{\pi}{2}\}$$

Work Step by Step

$$\sin\frac{x}{2}-\cos\frac{x}{2}=0$$ over the interval $[0,2\pi)$ 1) Solve the equation: All the problems in this exercise follows the method of solving portrayed in Example 4: SQUARING each side to transform the equation into one trigonometric function. $$\Big(\sin\frac{x}{2}-\cos\frac{x}{2}\Big)^2=0^2$$ $$\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)-2\sin\frac{x}{2}\cos\frac{x}{2}=0$$ - $\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)$ can be written into $1$, as $\sin^2A+\cos^2A=1$ - $\Big(2\sin\frac{x}{2}\cos\frac{x}{2}\Big)$ can be written into $\sin x$, as $2\sin A\cos A=\sin2A$ Therefore, $$1-\sin x=0$$ $$\sin x=1$$ Over the interval $[0, 2\pi)$, there is one value of $x$ where $\sin x=1$, which is $\frac{\pi}{2}$. Therefore, $$x=\{\frac{\pi}{2}\}$$ 2) Substitute the found solutions back to the original equation for checking (compulsory after using the squaring method) - For $x=\frac{\pi}{2}$: $$\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=0$$ Therefore, $\frac{\pi}{2}$ is a correct solution to the equation. In other words, the solution set is $$\{\frac{\pi}{2}\}$$
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