## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{0^\circ,90^\circ, 180^\circ, 270^\circ\}$$
$$\sin^2\theta\cos^2\theta=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$\sin^2\theta\cos^2\theta=0$$ $$(\sin\theta\cos\theta)^2=0$$ $$\sin\theta\cos\theta=0$$ $$\sin\theta=0\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=0$$ 2) Apply the inverse function: - For $\sin\theta=0$: Over the interval $[0^\circ,360^\circ)$, examining the unit circle, we easily find that there are 2 values of $\theta$ where $\sin\theta=0$, which are $\{0^\circ,180^\circ\}$ - For $\cos\theta=0$: Over the interval $[0^\circ,360^\circ)$, examining the unit circle, we also easily find that there are 2 values of $\theta$ where $\cos\theta=0$, which are $\{90^\circ,270^\circ\}$ Therefore, $$\theta\in\{0^\circ,90^\circ, 180^\circ, 270^\circ\}$$ In other words, the solution set to this problem is $$\{0^\circ,90^\circ, 180^\circ, 270^\circ\}$$