Question

cos4x=3- 2cos2x

Answer 1

\(cos4x+2cos2x=3\Leftrightarrow2cos^22x-1+2cos2x=3\)

\(\Leftrightarrow2cos^22x+2cos2x-4=0\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=cos\left(k2pi\right)\\cos2x=-2\left(loại\right)\end{matrix}\right.\Rightarrow x=kpi\)