2x = \frac{\pi}{2} + \pi n, n \in Z => x = \frac{\pi}{4} + \frac{\pi}{2} n, n \in Z" alt="sin^4x + cos^4x = sin2x - \frac{1}{2}\\ sin^4x + cos^4x + 2sin^2xcos^2x - 2sin^2xcos^2x = sin2x - \frac{1}{2}\\(sin^2x + cos^2x)^2 - \frac{1}{2}sin^22x = sin2x - \frac{1}{2}\\ 1 - \frac{1}{2}sin^22x = sin2x - \frac{1}{2}\\\frac{1}{2}sin^22x + sin2x - \frac{3}{2} = 0\\ sin^22x + 2sin2x - 3 = 0\\sin^22x + 2sin2x + 1 - 4 = 0\\(sin2x+1)^2 - 2^2 = 0\\(sin2x+3)(sin2x-1)=0\\sin2x = 1 => 2x = \frac{\pi}{2} + \pi n, n \in Z => x = \frac{\pi}{4} + \frac{\pi}{2} n, n \in Z" align="absmiddle" class="latex-formula">