## Trigonometry (11th Edition) Clone

The solution set is $$\{0^\circ+360^\circ n, 180^\circ+360^\circ n, n\in Z\}$$
$$\sin\theta\cos\theta-\sin\theta=0$$ 1) Solve the equation over the interval $[0^\circ,360^\circ)$ $$\sin\theta\cos\theta-\sin\theta=0$$ $$\sin\theta(\cos\theta-1)=0$$ $$\sin\theta=0\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=1$$ - For $\cos\theta=1$ Over the interval $[0^\circ, 360^\circ)$, there is only one value of $\theta$ where $\cos\theta=1$, which is $0^\circ$. - For $\sin\theta=0$ Over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\sin\theta=0$, which are $0^\circ$ and $180^\circ$. Therefore, overall, $$\theta=\{0^\circ, 180^\circ\}$$ (though $\theta=0^\circ$ appears in both cases, we only need to mention one time, since they are eventually the same point in the unit circle) 2) Solve the equation for all solutions - The integer multiples of the period of both the sine and cosine function is $360^\circ$. - We apply it to each solution found in step 1. So the solution set is $$\{0^\circ+360^\circ n, 180^\circ+360^\circ n, n\in Z\}$$