Question

[ngoclanne]

giúp em với ạ em cần gấp

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow-4sinx.cosx\left(cos^2x-sin^2x\right)-\sqrt{3}cos4x+1=0\)

\(\Leftrightarrow-2sin2x.cos2x-\sqrt{3}cos4x+1=0\)

\(\Leftrightarrow-sin2x-\sqrt{3}cos2x+1=0\)

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

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)