Question

Solve : $sin2x=\sqrt{2}cosx$

• A. $x=n\pi +\frac{\pi }{2}$
• B. $x=2n\pi +\pi /4$
• C. $x=2n\pi -\pi /4$
• D. $x=n\pi -\frac{\pi }{2}(n\in I)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $x=n\pi +\frac{\pi }{2}$
B $x=2n\pi +\pi /4$
D $x=n\pi -\frac{\pi }{2}(n\in I)$

$\sin 2x=\sqrt{2}\cos x$

$\Rightarrow -\cos x(\sqrt{2}-2\sin x)=0$

$\Rightarrow \cos x=0or(\sqrt{2}-2\sin x)=0$

$\Rightarrow \cos x=0$ or $\sin x=\frac{1}{\sqrt{2}}$

$\Rightarrow x=n\pi \pm \frac{\pi }{2},2n\pi +\frac{\pi }{4}$