Question

If the equation $4\sin (x+\frac{\pi }{3})\cos (x-\frac{\pi }{6})=a^2+\sqrt{3}\sin 2x-\cos 2x$ has a solution, then the value of $a$ can be

• A. $-3$
• B. $0$
• C. $2$
• D. $3$

Answer 1

Answer: (B), (C)

$2$

$4\sin (x+\frac{\pi }{3})\cos (x-\frac{\pi }{6})=a^2+\sqrt{3}\sin 2x-\cos 2x$
$\Rightarrow 4(\frac{\sin x}{2}+\frac{\sqrt{3}\cos x}{2})(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x)=a^2+\sqrt{3}\sin 2x-\cos 2x\Rightarrow (\sin x+\sqrt{3}\cos x{)}^2=a^2+\sqrt{3}\sin 2x-\cos 2x\Rightarrow 3{\cos }^{2}x+{\sin }^{2}x+\sqrt{3}\sin 2x=a^2+\sqrt{3}\sin 2x-\cos 2x\Rightarrow 2{\cos }^{2}x+1=a^2-\cos 2x\Rightarrow \cos 2x+1+1=a^2-\cos 2x\Rightarrow \cos 2x=\frac{a^2-2}{2}$
Since $\cos 2x\in [-1,1]$
$\therefore -1\le \frac{a^2-2}{2}\le 1$
$\Rightarrow 0\le a^2\le 4\Rightarrow -2\le a\le 2$