\(\left(cos^2x-sin^2x\right)\left(cos^4x+sin^4x+sin^2x.cos^2x\right)=cos2x\)
\(\Leftrightarrow cos2x\left(cos^4x+sin^4x+sin^2x.cos^2x\right)-cos2x=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\cos^4x+sin^4x+sin^2x.cos^2x-1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(sin^2x+cos^2x\right)^2-sin^2x.cos^2x-1=0\)
\(\Leftrightarrow sin^2x.cos^2x=0\)
\(\Leftrightarrow sin2x=0\Rightarrow x=\frac{k\pi}{2}\)
Tổng hợp lại ta được: \(x=\frac{k\pi}{4}\)
b/
\(\Leftrightarrow\frac{cos3x\left(cos3x+3cosx\right)}{4}+\frac{sin3x\left(3sinx-sin3x\right)}{4}-\frac{5}{8}=0\)
\(\Leftrightarrow cos^23x+3cosx.cos3x+3sin3x.sin3x-sin^23x-\frac{5}{2}=0\)
\(\Leftrightarrow\left(cos^23x-sin^23x\right)+3\left(cos3x.cosx+sin3x.sinx\right)-\frac{5}{2}=0\)
\(\Leftrightarrow cos6x+3cos2x-\frac{5}{2}=0\)
\(\Leftrightarrow4cos^32x-3cos2x+3cos2x-\frac{5}{2}=0\)
\(\Leftrightarrow cos^32x=\frac{5}{8}\)
\(\Rightarrow cos2x=\frac{\sqrt[3]{5}}{2}\)
\(\Rightarrow x=\pm\frac{1}{2}arccos\left(\frac{\sqrt[3]{5}}{2}\right)+k\pi\)
