Answer
$$\cos\theta=-\frac{\sqrt3}{2}$$
Work Step by Step
$$\cos2\theta=\frac{1}{2}\hspace{1cm}\theta\hspace{0.2cm}\text{terminates in quadrant II}\hspace{1cm}\cos\theta=?$$
From the original half-angle identity for cosines:
$$\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}$$
we can rewrite to use to find $\cos\theta$
$$\cos\theta=\pm\sqrt{\frac{1+\cos2\theta}{2}}$$
Also, whether to choose positive or negative square root now depends on the sign of $\cos\theta$.
$\theta$ terminates in quadrant II, where cosines are negative. Therefore, $\cos\theta\lt0$.
As a result, we need to choose negative square root:
$$\cos\theta=-\sqrt{\frac{1+\cos2\theta}{2}}$$
Now we can start calculating $\cos\theta$.
$$\cos\theta=-\sqrt{\frac{1+\frac{1}{2}}{2}}$$
$$\cos\theta=-\sqrt{\frac{\frac{3}{2}}{2}}$$
$$\cos\theta=-\sqrt{\frac{3}{4}}$$
$$\cos\theta=-\frac{\sqrt3}{2}$$
Therefore, $$\cos\theta=-\frac{\sqrt3}{2}$$
