Question

$sin^{2}7x-cos^{2}7x= \frac{ \sqrt{3}}{2}$

Answer 1

$\cos2 \alpha =\cos^2 \alpha -\sin^2 \alpha$

$\sin^{2}7x-\cos^{2}7x= \frac{ \sqrt{3}}{2} \\\ -\cos14x= \frac{ \sqrt{3}}{2} \\\ \cos14x= -\frac{ \sqrt{3}}{2} \\\ 14x=\pm \frac{5\pi }{6} +2 \pi n \\\ x=\pm \frac{5 \pi }{84} + \frac{ \pi n}{7} , \ n\in Z$