# Chapter 6 - Section 6.2 - More on Trigonometric Equations - 6.2 Problem Set - Page 332: 32

$\theta=\{90^o,210^o\}$

#### Work Step by Step

$sin(\theta)-\sqrt{3}cos(\theta)=1$ $sin(\theta)-1=\sqrt{3}cos(\theta)$ $[sin(\theta)-1]^2=[\sqrt{3}cos(\theta)]^2$ $sin^2(\theta)-2sin(\theta)+1=3cos^2(\theta)$ $sin^2(\theta)-2sin(\theta)+1=3-3sin^2(\theta)$ $4sin^2(\theta)-2sin(\theta)-2=0$ $2sin^2(\theta)-sin(\theta)-1=0$ $sin(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-1) \pm \sqrt{(-1)^2 - 4 (2)(-1)}}{2(2)}=1,\frac{-1}{2}$ $sin(\theta)=1$ $\theta=sin^{-1}(1)$ $\theta=90^o$ $sin(\theta)=\frac{-1}{2}$ $\theta=sin{-1}(-\frac{1}{2})$ We know $sin(\theta)$ is negative in quadrant $III$ and $IV$ $\theta=180^o+30^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360-30^o$ $\theta=210^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=330^o$ Refuse $\theta=\{90^o,210^o\}$

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