# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 22

The solution set to this problem is $$\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$

#### Work Step by Step

$$(\csc x+2)(\csc x-\sqrt2)=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$(\csc x+2)(\csc x-\sqrt2)=0$$ $$\csc x+2=0\hspace{1cm}\text{or}\hspace{1cm}\csc x-\sqrt2=0$$ $$\csc x=-2\hspace{1cm}\text{or}\hspace{1cm}\csc x=\sqrt2$$ 2) Apply the inverse function: - $\csc x=-2$: $\csc\lt0$ means that the angle of $x$ lies either in quadrant III or IV. In quadrant I, $\csc x=2$ refers to angle $\frac{\pi}{6}$; therefore, in quadrant III and IV, $\csc x=-2$ would refer to angle $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ - $\csc x=\sqrt2$: $\csc\gt0$ means that the angle of $x$ lies either in quadrant I or II. In quadrant I, $\csc x=\sqrt2$ refers to angle $\frac{\pi}{4}$; in quadrant II, it refers to angle $\frac{3\pi}{4}$ Therefore, $$x\in\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$ In other words, the solution set to this problem is $$\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$

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