Question

Solve for x, $(-\pi \le x\le \pi )$ the equation : $2(\cos x+\cos 2x)+\sin 2x(1+2\cos x)=2\sin x$.
How many distinct values of x satisfy the equation in the above range

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

Given, $2(\cos x+\cos 2x)+\sin 2x(1+2\cos x)=2\sin x,for(-\pi \le x\le \pi ),2(\cos x+2{\cos }^{2}x-1)+2\sin x\cos x(1+2\cos x)=2\sin x,⟹(\cos x+2{\cos }^{2}x-1)+\sin x(\cos x+2{\cos }^{2}x-1)=0⟹(\cos x+2{\cos }^{2}x-1)(1+\sin x)=0,\sin x=-1,xϵ\{-\frac{\pi }{2}\}\cos x+2{\cos }^{2}x-1,(2\cos x-1)(\cos x+1),xϵ\{-\pi ,-\frac{\pi }{3},\frac{\pi }{3},\pi \}i.e,5solutions,xϵ\{-\pi ,-\frac{\pi }{2},-\frac{\pi }{3},\frac{\pi }{3},\pi \}$