Question

$2 \sin^{2} (2x) - 1 = 0$
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Answer 1

$2sin^22x-1=0\\\\sin^22x=\frac{1}{2}\; \; ,\qquad \Big [\; sin^2a=\frac{1-cos2a}{2}\; \Big ]\\\\\frac{1-cos4x}{2}=\frac{1}{2}\\\\1-cos4x=1\\\\cos4x=0\\\\4x=\frac{\pi }{2}+\pi n\; ,\; n\in Z\\\\x=\frac{\pi }{8}+\frac{\pi n}{4}\; ,\; n\in Z$