## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{0^\circ,30^\circ, 150^\circ, 180^\circ\}$$
$$2\tan^2\theta\sin\theta-\tan^2\theta=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$2\tan^2\theta\sin\theta-\tan^2\theta=0$$ $$\tan^2\theta(2\sin\theta-1)=0$$ $$\tan^2\theta=0\hspace{1cm}\text{or}\hspace{1cm}2\sin\theta-1=0$$ $$\tan\theta=0\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=\frac{1}{2}$$ 2) Apply the inverse function: - For $\tan\theta=0$: Over the interval $[0^\circ,360^\circ)$, there are 2 values of $\theta$ where $\tan\theta=0$, which are $\{0^\circ,180^\circ\}$ - For $\sin\theta=\frac{1}{2}$: Over the interval $[0^\circ,360^\circ)$, in quadrant I, we already know that as $\theta=30^\circ$, $\sin\theta=\frac{1}{2}$. However, in quadrant II, as $\theta=150^\circ$, $\sin\theta$ also equals $\frac{1}{2}$. Therefore, $$\theta\in\{0^\circ,30^\circ, 150^\circ, 180^\circ\}$$ In other words, the solution set to this problem is $$\{0^\circ,30^\circ, 150^\circ, 180^\circ\}$$