Решите пожалуйста подробно ДАЮ 40 БАЛЛОВ! sin^6x+cos^6x=1/4

Решите задачу:

$sin^6x+cos^6x=\frac{1}{4}\\\\\\\star \; \; sin^2x+cosx^2=1\; \; \to \; \; \; 1=1^3=(sin^2x+cos^2x)^3\; \; \to \\\\(sin^2x+cos^2x)^3=sin^6x+3sin^4x\cdot cos^2x+3sin^2x\cdot cos^4x+cos^6x=\\\\=sin^6x+cos^6x+3sin^2x\cdot cos^2x(\underbrace {sin^2x+cos^2x}_{1})\; \; \; \; \to\\\\1=sin^6x+cos^6x+3sin^2x\cdot cos^2x\; \; \; \to \\\\sin^6x+cos^6x=1-3(\underbrace {sinx\cdot cosx}_{\frac{1}{2}sin2x})^2\\\\sin^6x+cos^6x=1-3\cdot \frac{1}{4}sin^22x\; \; \; \star \\\\\\1-\frac{3}{4}sin^22x=\frac{1}{4}$

$\frac{3}{4}sin^22x=\frac{3}{4}\\\\sin^22x=1\; \; \quad , \; \; \star \; \; \; sin^2 \alpha =\frac{1-cos2 \alpha }{2}\; \; \; \star \\\\\frac{1-cos4x}{2}=1\; ,\; \; 1-cos4x=2\\\\ cos4x=-1\\\\4x=\pi +2\pi n,\; n\in Z\\\\x=\frac{\pi}{4}+\frac{\pi n}{2}\; ,\; n\in Z$