Question

The equation $4\sin (x+\frac{\pi }{3})\cos (x-\frac{\pi }{6})=a^2+\sqrt{3}\sin 2x-\cos 2x$ has a solution if the value of a is

• A. $-2$
• B. $0$
• C. $2$
• D. $a\in [-2,2]$

Answer 1

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

The correct options are
A $2$
B $0$
C $a\in [-2,2]$
D $-2$

Given equation $4\sin (x+\frac{\pi }{3})\cos (x-\frac{\pi }{6})=a^2+\sqrt{3}\sin 2x-\cos 2x$
$4[\sin x\cos \frac{\pi }{3}+\cos x\sin \frac{\pi }{3}]\times [\cos x\cos \frac{\pi }{6}+\sin x\sin \frac{\pi }{6}]=a^2+\sqrt{3}\sin 2x-\cos 2x$
$\Rightarrow 4[\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x][\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x]=a^2+\sqrt{3}\sin 2x-\cos 2x$
$\Rightarrow \sqrt{3}\sin 2x+3{\cos }^{2}x+{\sin }^{2}x=a^2+\sqrt{3}\sin 2x-\cos 2x$
$\Rightarrow \cos 2x+2=a^2-\cos 2x$
$\Rightarrow \cos 2x=\frac{a^2-2}{2}$
$\Rightarrow -1\le \frac{a^2-2}{2}\le 1$
$\Rightarrow 0\le a^2\le 4$
$\Rightarrow -2\le a\le 2$.
all values of a given in (a), (b), (c), (d) satisfy this relation.