Question

\(cos^2x+cos^22x+cos^23x+cos^24x=\frac{3}{2}\)

Answer 1

\(\Leftrightarrow2cos^2x-1+2cos^22x-1+2cos^23x-1+2cos^24x=0\)

\(\Leftrightarrow cos2x+cos4x+cos6x+2cos^24x=0\)

\(\Leftrightarrow2cos4x.cos2x+cos4x+2cos^24x=0\)

\(\Leftrightarrow cos4x\left(2cos2x+1+2cos4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2cos4x+2cos2x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(2cos^22x-1\right)+2cos2x+1=0\)

\(\Leftrightarrow4cos^22x+2cos2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2x=\frac{\sqrt{5}-1}{4}=cos\left(\frac{2\pi}{5}\right)\\cos2x=\frac{-\sqrt{5}-1}{4}=cos\left(\frac{4\pi}{5}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)