Question

Solve for $x,(-\pi \le x\le \pi )$, the equation $2(cosx+cos2x)+sin2x(1+2cosx)=2sinx$

• A. $\pm \pi$
• B. $\pm \frac{\pi }{2}$
• C. $\pm \frac{\pi }{3}$
• D. None of these

Answer 1

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

The correct options are
A $\pm \pi$
B $\pm \frac{\pi }{2}$
C $\pm \frac{\pi }{3}$

$2\cos x+2\cos 2x+\sin 2x+\sin 3x+\sin x-2\sin x=0\therefore 2\cos x+2\cos 2x+2\sin x\cos x+(\sin 3x-\sin x)=0\Rightarrow 2\cos x+2\cos 2x+2\sin x\cos x+2\cos 2x\sin x=0\Rightarrow 2\cos x(1+\sin x)+2\cos 2x(1+\sin x)=0\Rightarrow 2(1+\sin x)(\cos x+\cos 2x)=0$
$\Rightarrow 4(1+\sin x)\cos \left(\frac{3x}{2}\right)\cos \left(\frac{x}{2}\right)=0$
$\Rightarrow \sin x=-1$ or $\cos \left(\frac{3x}{2}\right)=0$ or $\cos \left(\frac{x}{2}\right)=0$
$\Rightarrow x=2n\pi +\frac{3\pi }{2}$ or $x=(2n+1)\frac{\pi }{3}$ or $x=(2n+1)\pi$
For x belong to $[-\pi ,\pi ]$
$x=-\pi ,-\frac{\pi }{2},-\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{3},\pi$