## Precalculus (6th Edition) Blitzer

The solution is $\underline{x=\frac{\pi }{6},\frac{11\pi }{6}}.$
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 $\left[ 0,2\pi \right).$