## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\}$$
$$-2\sin^2x=3\sin x+1$$ over interval $[0,2\pi)$ 1) Solve the equation: $$-2\sin^2x=3\sin x+1$$ $$2\sin^2x+3\sin x+1=0$$ $$(2\sin^2x+2\sin x)+(\sin x+1)=0$$ $$(\sin x+1)(2\sin x+1)=0$$ $$\sin x=-1\hspace{1cm}\text{or}\hspace{1cm}\sin x=-\frac{1}{2}$$ 2) Apply the inverse function: $\sin x=-1$: Over the interval $[0,2\pi)$, looking at the unit circle, we see that there is only 1 value where sine equals $-1$, which is $\frac{3\pi}{2}$. $\sin x=-\frac{1}{2}$: Over the interval $[0,2\pi)$, $\sin x\lt0$ means the value of $x$ lies either in quadrant III or IV. The reference angle is $\frac{\pi}{6}$. Thus, $x=\frac{7\pi}{6}$ in quadrant III and $x=\frac{11\pi}{6}$ in quadrant IV, either of which would get $\sin x=-\frac{1}{2}$ Therefore, $$x\in\{\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\}$$ In other words, the solution set to this problem is $$\{\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\}$$