## Trigonometry (11th Edition) Clone

The solution set is $$\{90^\circ+360^\circ n, 221.8^\circ+360^\circ n, 318.2^\circ+360^\circ n, n\in Z\}$$
$$3\sin^2\theta-\sin\theta=2$$ 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 $$\{90^\circ+360^\circ n, 221.8^\circ+360^\circ n, 318.2^\circ+360^\circ n, n\in Z\}$$