Question

Tìm min: \(y=sin^4x+cos^4x+sinx.cosx\)

Answer 1

\(y=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x+sinx.cosx\)

\(y=-\frac{1}{2}sin^22x+\frac{1}{2}sin2x+1\)

\(y=-\frac{1}{2}\left(sin2x-\frac{1}{2}\right)^2+\frac{9}{8}\ge\frac{9}{8}\)

\(y_{min}=\frac{9}{8}\) khi \(sin2x=\frac{1}{2}\)