Answer
$$\cos x=-\frac{\sqrt3}{2}$$
$$\tan x=\frac{\sqrt3}{3}$$
Work Step by Step
$$\sin x=-\frac{1}{2}, x\in\Big[\pi,\frac{3\pi}{2}\Big]$$
1) As $x\in\Big[\pi,\frac{3\pi}{2}\Big]$, it means angle $x$ is in the third quadrant, so $\cos x\lt0$ and $\tan x\gt0$.
2) To find $\cos x$, we employ the formula: $$\cos^2x=1-\sin^2x=1-\Big(-\frac{1}{2}\Big)^2=1-\frac{1}{4}=\frac{3}{4}$$
$$|\cos x|=\frac{\sqrt3}{2}$$
Since $\cos x\lt0$, $$\cos x=-\frac{\sqrt3}{2}$$
3) To find $\tan x$, we employ the formula:
$$\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$$
