Giải phương trình: sin2x−3√2sin2x+2cos2x=1

Question

Giải phương trình:

 \(\sin ^{2} x-\frac{\sqrt{3} }{2} \sin 2x+2\cos ^{2} x=1.  \)

Answer 1

\(\sin ^{2} x-\frac{\sqrt{3} }{2} \sin 2x+2\cos ^{2} x=1.      \)            
\(\sin ^{2} x-\frac{\sqrt{3} }{2} \sin 2x+2\cos ^{2} x=1\Leftrightarrow \frac{1-\cos 2x}{2} -\frac{\sqrt{3} }{2} \sin 2x+2.\frac{1+\cos 2x}{2} =1 \)
\(\Leftrightarrow \frac{\sqrt{3} }{2} \sin 2x-\frac{1}{2} \cos 2x=\frac{1}{2} \Leftrightarrow \sin \left(2x-\frac{\pi }{6} \right)=\frac{1}{2} \) 
\(\Leftrightarrow \left[\begin{array}{l} {2x-\frac{\pi }{6} =\frac{\pi }{6} +k2\pi } \\ {2x-\frac{\pi }{6} =\pi -\frac{\pi }{6} +k2\pi } \end{array}\right. \, \left(k\in {\rm Z}\right)\Leftrightarrow \left[\begin{array}{l} {x=\frac{\pi }{6} +k\pi } \\ {x=\frac{\pi }{2} +k\pi } \end{array}\right. \, \, \, \left(k\in {\rm Z}\right).  \)