Question

The number of solutions of the equation $\cos 3x+\cos 2x=\sin \frac{3x}{2}+\sin \frac{x}{2}$ lying in the interval $[0,2\pi ]$ is

• A. $5$
• B. $6$
• C. $7$
• D. $4$

Answer 1

The correct option is A $5$

$\cos 3x+\cos 2x=\sin \frac{3x}{2}+\sin \frac{x}{2}$
$\Rightarrow 2\cos \frac{5x}{2}\cos \frac{x}{2}=2\sin x\cos \frac{x}{2}$
$\Rightarrow \cos \frac{x}{2}(\cos \frac{5x}{2}-\sin x)=0$
$\Rightarrow \cos \frac{x}{2}=0$ or $\cos \frac{5x}{2}=\sin x=\cos (\frac{\pi }{2}-x)$

$\Rightarrow \frac{x}{2}=(2n+1)\frac{\pi }{2}\Rightarrow x=(2n+1)\pi$
or $\frac{5x}{2}=2n\pi \pm (\frac{\pi }{2}-x)$

Taking positive sign
$\frac{7x}{2}=2n\pi +\frac{\pi }{2}\Rightarrow x=(4n+1)\frac{\pi }{7}$

Taking negative sign
$\frac{3x}{2}=2n\pi -\frac{\pi }{2}$
$\Rightarrow x=(4n-1)\frac{\pi }{3}$
Since $0\le x\le 2\pi ,$
$x=\pi ,\frac{\pi }{7},\frac{5\pi }{7},\frac{9\pi }{7},\frac{13\pi }{7}$