Question

giải phương trình sau

Sin2x -\(\sqrt{3}\) cos 2x =\(\sqrt{2}\)

 

Answer 1

\(sin2x-\sqrt{3}cos2x=\sqrt{2}\)

\(\Rightarrow2sin\left(2x-\dfrac{\pi}{3}\right)=\sqrt{2}\)

\(\Rightarrow sin\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}=sin\dfrac{\pi}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7\pi}{24}+k\pi\\x=\dfrac{\pi}{24}+k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)