Question

y= (căn3).cos2x - sin2x + 2x.

Giải phương trình y'=0

Answer 1

\(y'=-2\sqrt{3}sin2x-2cos2x+2=0\)

\(\Leftrightarrow\sqrt{3}sin2x+cos2x=1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=\frac{1}{2}\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\2x+\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)