\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sin4x=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{2}sin4x=0\)
\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)+\frac{1}{2}sin4x=0\)
\(\Leftrightarrow3cos4x+4sin4x=-5\)
\(\Leftrightarrow\frac{4}{5}sin4x+\frac{3}{5}cos4x=-1\)
Đặt \(\frac{4}{5}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sin\left(4x+a\right)=-1\)
\(\Leftrightarrow4x+a=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{1}{4}a-\frac{\pi}{8}+\frac{k\pi}{2}\)