Find value of cos(2x)

Given
$\sin x=\frac14$ and $x\in(\pi/2,\pi)$ so cosine will be negative in second quadrant
Thus,
$$\cos x=-\sqrt{1-\sin^2x}=-\sqrt{1-\frac1{16}}=-\frac{\sqrt15}{4}$$
Thus,
$$\cos2x=\cos^2x-\sin^2x=(-\frac{\sqrt15}{4})^2-(\frac14)^2=\frac{15}{16}-\frac1{16}=\boxed{\frac78}$$