Question

If $(2\sin 3x\cos x-2\cos 3x\sin x{)}^2=2,$ then $x$ equals to

• A. $-\frac{\pi }{8}$
• B. $\frac{5\pi }{8}$
• C. $\frac{3\pi }{4}$
• D. $\frac{3\pi }{8}$

Answer 1

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

The correct options are
A $-\frac{\pi }{8}$
B $\frac{5\pi }{8}$
D $\frac{3\pi }{8}$

$(2\sin 3x\cos x-2\cos 3x\sin x{)}^2=2$
$\Rightarrow \left(2\sin \right(3x-x){)}^2=2$
$\because \sin (A-B)=\sin A\cos B-\cos A\sin B$
$\Rightarrow {\sin }^2\left(2x\right)=\frac{1}{2}={\left(\frac{1}{\sqrt{2}}\right)}^2$
$\Rightarrow {\sin }^2\left(2x\right)={\sin }^2\left(\frac{\pi }{4}\right)$
$\Rightarrow 2x=n\pi \pm \frac{\pi }{4},n\in Z$
$\Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{8},n\in Z$