## Precalculus (6th Edition) Blitzer

$\frac{\pi }{6}$ is not the solution of the given equation.
We know that according to the ratio table of trigonometry, the value of $\cos \frac{\pi }{6}$ is given as $\frac{\sqrt{3}}{2}$, and $\sin \frac{\pi }{6}$ is given as $\frac{1}{2}$. \begin{align} & \cos x+2=\sqrt{3}\sin x \\ & \cos \frac{\pi }{6}+2=\sqrt{3}\sin \frac{\pi }{6} \\ & \frac{\sqrt{3}}{2}+2=\sqrt{3}.\frac{1}{2} \\ & \frac{\sqrt{3}+4}{2}=\frac{\sqrt{3}}{2} \end{align} Thus, $\cos x+2=\sqrt{3}\sin x$ is false.