## Precalculus (6th Edition) Blitzer

For the provided equation, the value of x is $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$.
We have the provided equation as $4{{\cos }^{2}}x=3$ Then, solving, \begin{align} & 4{{\cos }^{2}}x=3 \\ & {{\cos }^{2}}x=\frac{3}{4} \\ & \cos x=\pm \frac{\sqrt{3}}{2} \end{align} For $\cos x=\frac{\sqrt{3}}{2}$ in the given range, $x=\frac{\pi }{6},\frac{11\pi }{6}$ And for $\cos x=\frac{-\sqrt{3}}{2}$ $x=\frac{5\pi }{6},\frac{7\pi }{6}$ Hence, we have $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$.