I suppose x should be π.
Recall the double angle identity for cosine:
[tex]\cos^2\dfrac x2=\dfrac{1+\cos x}2[/tex]
Then remember for [tex]0<x<\frac\pi2[/tex], we have [tex]\cos x>0[/tex].
Let [tex]x=\frac{7\pi}6[/tex]. Plugging this into the equation above gives
[tex]\cos^2\dfrac{7\pi}{12}=\dfrac{1+\cos\frac{7\pi}6}2[/tex]
Take the square root of both sides; this introduces two possible values, but we know [tex]\cos\frac{7\pi}{12}[/tex] should be positive, so
[tex]\cos\dfrac{7\pi}{12}=\sqrt{\dfrac{1+\cos\frac{7\pi}6}2}=\dfrac{\sqrt{2-\sqrt3}}2[/tex]
