Question

2√3 sin²3x - cos6x = √3 

Answer 1

\(2\sqrt{3}\sin^23x-cos6x=\sqrt{3}\\ \Leftrightarrow2sin^23x\left(\sqrt{3}+1\right)=\sqrt{3}+1\\ \Leftrightarrow sin^23x=\dfrac{1}{2}\Leftrightarrow\dfrac{1-cos6x}{2}=\dfrac{1}{2}\Rightarrow cos6x=0\Leftrightarrow x=\pm\dfrac{\pi}{12}+k\dfrac{\pi}{6}\left(k\in Z\right)\)