## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 279: 30

#### Answer

The solution set is $\{\varnothing\}$

#### Work Step by Step

$$\sin^2\frac{x}{2}-2=0$$ over interval $[0,2\pi)$ 1) Find corresponding interval for $\frac{x}{2}$ As the interval for $x$ is $[0,2\pi)$, we can only write it as the inequality $$0\le x\lt2\pi$$ Therefore, $$0\le\frac{x}{2}\lt\pi$$ So the interval for $\frac{x}{2}$ is $[0,\pi)$. 2) Now consider back the equation $$\sin^2\frac{x}{2}-2=0$$ $$\sin^2\frac{x}{2}=2$$ $$\sin\frac{x}{2}=\pm\sqrt2$$ Recall the defined range of $\sin\theta$ is $[-1,1]$. Here we need to find values of $\frac{x}{2}$ so that $\sin\frac{x}{2}=\pm\sqrt2$. However, as $\sqrt2\gt1$ and $-\sqrt2\lt(-1)$, it follows that $\pm\sqrt2\notin[-1,1]$, the range of sine. That means there is no value of $\frac{x}{2}$ that $\sin\frac{x}{2}$ could equal $\pm\sqrt2$ That leads to no values of $x$ which could satisfy the given equation. In other words, the solution set is $\{\varnothing\}$

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