## Precalculus (6th Edition) Blitzer

The solution of given equation is $\underline{\frac{\pi }{6},\frac{11\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6}.}$
We have to use the definition of the modulus function. That means \begin{align} & \left| y \right|=a\text{ then,} \\ & y=a\text{ or y}=-a \\ \end{align} So, for the provided function, \begin{align} & \left| \cos x \right|=\frac{\sqrt{3}}{2} \\ & \cos x=\frac{\sqrt{3}}{2} \\ \end{align} Or, $\cos x=-\frac{\sqrt{3}}{2}$ When $\cos x=\frac{\sqrt{3}}{2}$. On the provided interval $\left[ 0,2\pi \right)$, the cosine function is $\frac{\sqrt{3}}{2}$ at $\frac{\pi }{6}\text{ and }\frac{11\pi }{6}$. That means: \begin{align} & \cos x=\frac{\sqrt{3}}{2} \\ & x=\frac{\pi }{6},\frac{11\pi }{6} \end{align} When $\cos x=-\frac{\sqrt{3}}{2}$. On the provided interval $\left[ 0,2\pi \right)$, the cosine function is $-\frac{\sqrt{3}}{2}$ at $\frac{5\pi }{6}\text{ and }\frac{7\pi }{6}$. That means: \begin{align} & \cos x=-\frac{\sqrt{3}}{2} \\ & x=\frac{5\pi }{6},\frac{7\pi }{6} \end{align} Thus, the solution of $\left| \cos x \right|=\frac{\sqrt{3}}{2}$ on the interval $\left[ 0,2\pi \right)$ is $\underline{x=\frac{\pi }{6},\frac{11\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6}}.$