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: 52


The solution set is $$\{0,\pi\}$$

Work Step by Step

$$\sin\frac{x}{2}+\cos\frac{x}{2}=1$$ 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. $$\Big(\sin\frac{x}{2}+\cos\frac{x}{2}\Big)^2=1^2$$ $$\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)+2\sin\frac{x}{2}\cos\frac{x}{2}=1$$ - $\Big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\Big)$ 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, $$1+\sin x=1$$ $$\sin x=0$$ Over the interval $[0, 2\pi)$, there are two values of $x$ where $\sin x=0$, which are $0$ and $\pi$. Therefore, $$x=\{0,\pi\}$$ 2) Substitute the found solutions back to the original equation for checking (compulsory after using the squaring method) - For $x=0$: $$\sin0+\cos0=0+1=1$$ $x=0$ is a correct solution. - For $x=\pi$: $$\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1+0=1$$ $x=\pi$ is also a correct solution. In other words, the solution set is $$\{0,\pi\}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.