Answer
The solution of given equation is $\underline{\frac{\pi }{6},\frac{11\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6}.}$
Work Step by Step
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 $\underline{x=\frac{\pi }{6},\frac{11\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6}}.$
