Question

For which values of $a$ does the equation $4sin(x+\frac{\pi }{3})cos(x-\frac{\pi }{6})=a^2+\sqrt{3}sin2x-cos2x$ have solutions? Find the solutions for $a=0$, if exist.

• A. $-1\le a\le 1,$ $x=n\pi +\pi /2,$ $nϵz$
• B. $-2<a<2,$ $x=n\pi +\pi /2,$ $nϵz$
• C. $-2\le a\le 2,$ $x=n\pi +\pi /2,$ $nϵz$
• D. $-2\le a\le 2,$ $x=n\pi /2+\pi /2,$ $nϵz$

Answer 1

The correct option is C $-2\le a\le 2,$ $x=n\pi +\pi /2,$ $nϵz$

Gven, $4\sin (x+\frac{\pi }{3})\cos (x-\frac{\pi }{6})=a^2+\sqrt{3}\sin 2x-\cos 2x,$

$\sin (x+\frac{\pi }{3})=\sin (\frac{\pi }{2}+(x-\frac{\pi }{6}\left)\right)=\cos (x-\frac{\pi }{6}),$

$\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6},\frac{1}{2}=\sin \pi 6$

$\cos \frac{\pi }{6}\sin 2x-sin\frac{\pi }{6}\cos 2x=\sin (2x-\frac{\pi }{6}),$

$⟹4{\cos }^2(x-\frac{\pi }{6})=a^2+2\sin (2x-\frac{\pi }{6}),$

$2+2\cos (2x-\frac{\pi }{3})-2\sin (2x-\frac{\pi }{6})=a^2$

$-2<2\cos (2x-\frac{\pi }{3})-2\sin (2x-\frac{\pi }{6})<2$

$\therefore -2<a<2,$

$a=0,$

$-2=2\cos (2x-\frac{\pi }{3})-2\sin (2x-\frac{\pi }{6}),$

$\sin (2x-\frac{\pi }{6})-\cos (2x-\frac{\pi }{3})=1,$

$\sin (2x-\frac{\pi }{6})+\sin (2x-\frac{\pi }{6})=1$

$2x-\frac{\pi }{6}=2n\pi +\frac{\pi }{3}$

$x=n\pi +\frac{\pi }{2}$