Answer
The solution set is $$\{90^\circ+360^\circ n, 221.8^\circ+360^\circ n, 318.2^\circ+360^\circ n, n\in Z\}$$
Work Step by Step
$$3\sin^2\theta-\sin\theta=2$$
1) Solve the equation over the interval $[0^\circ,360^\circ)$
$$3\sin^2\theta-\sin\theta=2$$
$$3\sin^2\theta-\sin\theta-2=0$$
$$(3\sin^2\theta-3\sin\theta)+(2\sin\theta-2)=0$$
$$3\sin\theta(\sin\theta-1)+2(\sin\theta-1)=0$$
$$(\sin\theta-1)(3\sin\theta+2)=0$$
$$\sin\theta=1\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=-\frac{2}{3}$$
- For $\sin\theta=-\frac{2}{3}$: $$\theta=\sin^{-1}\Big(-\frac{2}{3}\Big)$$
$$\theta\approx221.8^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx318.2^\circ$$
- For $\sin\theta=1$
Over the interval $[0^\circ, 360^\circ)$, there is one value of $\theta$ where $\sin\theta=1$, which is $90^\circ$.
Therefore, overall, $$\theta=\{90^\circ, 221.8^\circ, 318.2^\circ\}$$
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 $$\{90^\circ+360^\circ n, 221.8^\circ+360^\circ n, 318.2^\circ+360^\circ n, n\in Z\}$$
