# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 51

The solution set is $$\{19.5^\circ+360^\circ n, 160.5^\circ+360^\circ n, 210^\circ+360^\circ n, 330^\circ+360^\circ n, n\in Z\}$$

#### Work Step by Step

$$6\sin^2\theta+\sin\theta=1$$ 1) Solve the equation over the interval $[0^\circ,360^\circ)$ $$6\sin^2\theta+\sin\theta=1$$ $$6\sin^2\theta+\sin\theta-1=0$$ $$(6\sin^2\theta-2\sin\theta)+(3\sin\theta-1)=0$$ $$2\sin\theta(3\sin\theta-1)+(3\sin\theta-1)=0$$ $$(3\sin\theta-1)(2\sin\theta+1)=0$$ $$\sin\theta=\frac{1}{3}\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=-\frac{1}{2}$$ - For $\sin\theta=\frac{1}{3}$: $$\theta=\sin^{-1}\frac{1}{3}$$ $$\theta\approx19.5^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx160.5^\circ$$ - For $\sin\theta=-\frac{1}{2}$ Over the interval $[0^\circ, 360^\circ)$, there are two values of $\theta$ where $\sin\theta=-\frac{1}{2}$, which are $210^\circ$ and $330^\circ$. Therefore, overall, $$\theta=\{19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the sine function is $360^\circ$. - We apply it to each solution found in step 1. So the solution set is $$\{19.5^\circ+360^\circ n, 160.5^\circ+360^\circ n, 210^\circ+360^\circ n, 330^\circ+360^\circ n, n\in Z\}$$

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