## Precalculus (6th Edition) Blitzer

The solutions for the equation are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{4\pi }{4},\frac{2\pi }{3},\frac{5\pi }{3}$.
\begin{align} & \cos x\cdot {{\tan }^{2}}x=3\cos x \\ & \cos x\cdot {{\tan }^{2}}x-3\cos =0 \\ & \cos x\left( {{\tan }^{2}}x-3 \right)=0 \end{align} If $\cos x=0$ Then, $x=\frac{\pi }{2},\frac{3\pi }{2}$ \begin{align} & {{\tan }^{2}}x-3=0 \\ & {{\tan }^{2}}x=3 \end{align} \begin{align} & \tan x=\pm \sqrt{3} \\ & \tan x=\sqrt{3} \\ \end{align} $x=\frac{\pi }{3},\frac{4\pi }{4}$ $\tan x=-\sqrt{3}$ $\therefore x=\frac{2\pi }{3},\frac{5\pi }{3}$ Hence, the solutions for the equation are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{4\pi }{4},\frac{2\pi }{3},\frac{5\pi }{3}$.