Question

Cos^2x -sin^2x= \|3/2

Answer 1

$cos^2x -sin^2x=\frac{\sqrt{3}}{2}$
$cos^2x - (1-cos^2x) = \frac{ \sqrt{3} }{2}$
$cos^2x - 1+cos^2x = \frac{ \sqrt{3} }{2}$
$2cos^2x -1 = \frac{ \sqrt{3} }{2}$
$cos2x = \frac{ \sqrt{3} }{2}$
$2x = arccos\frac{ \sqrt{3} }{2}$
$2x = 30$°
$x = 15$°