## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{\frac{\pi}{6},\frac{\pi}{2},\frac{3\pi}{2},\frac{11\pi}{6}\}$$
$$2\cos^2x-\sqrt3\cos x=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$2\cos^2x-\sqrt3\cos x=0$$ $$\cos x(2\cos x-\sqrt3)=0$$ $$\cos x=0\hspace{1cm}\text{or}\hspace{1cm}\cos x=\frac{\sqrt3}{2}$$ 2) Apply the inverse function: $\cos x=0$: Over the interval $[0,2\pi)$, looking at the unit circle, we see that there are 2 values where cosine equals $0$, which are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ $\cos x=\frac{\sqrt3}{2}$: Over the interval $[0,2\pi)$, $\cos x\gt0$ means the value of $x$ lies either in quadrant I or IV. In fact, $x=\frac{\pi}{6}$ in quadrant I and $x=\frac{11\pi}{6}$ in quadrant IV, either of which would get $\cos x=\frac{\sqrt3}{2}$ Therefore, $$x\in\{\frac{\pi}{6},\frac{\pi}{2},\frac{3\pi}{2},\frac{11\pi}{6}\}$$ In other words, the solution set to this problem is $$\{\frac{\pi}{6},\frac{\pi}{2},\frac{3\pi}{2},\frac{11\pi}{6}\}$$