Answer
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 $$\{19.5^\circ+360^\circ n, 160.5^\circ+360^\circ n, 210^\circ+360^\circ n, 330^\circ+360^\circ n, n\in Z\}$$
