Answer
The solution is $\underline{x=\frac{\pi }{6},\frac{11\pi }{6}}.$
Work Step by Step
The expression can be solved as follows:
$3\cos x-6\sqrt{3}=\cos x-5\sqrt{3}$
Arrange the provided equation in such a way that the cosine functions are on one side and the real numbers are on the other side:
$\begin{align}
& 3\cos x-\cos x=6\sqrt{3}-5\sqrt{3} \\
& 2\cos x=\sqrt{3} \\
& \cos x=\frac{\sqrt{3}}{2}
\end{align}$
So, in the interval $\left[ 0,2\pi \right),$ the cosine function is $\frac{\sqrt{3}}{2}$ at $\frac{\pi }{6}\text{ and }\frac{11\pi }{6}$. So,
$\begin{align}
& \cos x=\frac{\sqrt{3}}{2} \\
& x=\frac{\pi }{6}
\end{align}$
And the other value is:
$\begin{align}
& \cos x=\frac{\sqrt{3}}{2} \\
& =2\pi -\frac{\pi }{6} \\
& =\frac{11\pi }{6}
\end{align}$
And the value of cos is positive in the fourth quadrant.
The period of the sine function is $2\pi $, so the general solution of the equation is:
$x=\frac{\pi }{6}+2n\pi $
The other value is:
$x=\frac{11\pi }{6}+2n\pi $
Now, to get different solutions, $\text{put }n=0,1,2,3\ldots $
$\text{When }n=0$
$\begin{align}
& x=\frac{\pi }{6}+2n\pi \\
& =\frac{\pi }{6}
\end{align}$
And the other solution will be:
$\begin{align}
& x=\frac{11\pi }{6}+2n\pi \\
& =\frac{11\pi }{6}
\end{align}$
When we put other values of $n$, the solution becomes outside the interval $\left[ 0,2\pi \right).$
