Question

Giải phương trình:

\(2sin^2\dfrac{x}{2}-3cos\dfrac{x}{4}=0\)

Answer 1

\(\Leftrightarrow2cos2x.cos\left(\dfrac{\pi}{6}\right)-2sin2x.sin\left(\dfrac{\pi}{6}\right)+2sin2x-1=0\)

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

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

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

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

\(\Leftrightarrow...\)