Question

\(\sin^8x+\cos^8x=\frac{17}{32}\)

Answer 1

\(\Leftrightarrow\left(sin^4x+cos^4x\right)^2-2sin^4x.cos^4x=\frac{17}{32}\)

\(\Leftrightarrow\left[1-2sin^2x.cos^2x\right]^2-2sin^4x.cos^4x=\frac{17}{32}\)

Đặt \(sin^2x.cos^2x=\frac{1}{4}sin^22x=t\Rightarrow0\le t\le\frac{1}{4}\)

\(\Rightarrow\left(1-2t\right)^2-2t^2=\frac{17}{32}\)

\(\Leftrightarrow2t^2-4t+\frac{15}{32}=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{15}{8}\left(l\right)\\t=\frac{1}{8}\end{matrix}\right.\) \(\Rightarrow\frac{1}{4}sin^22x=\frac{1}{8}\Leftrightarrow2sin^22x=1\)

\(\Leftrightarrow cos4x=0\)