Question

Solve $\cos 3x+\cos 2x=\sin \left(3x/2\right)+\sin \left(x/2\right),0\le x<2\pi$.

Answer 1

$\cos 3x+\cos 2x=\sin \left(3x/2\right)+\sin \left(x/2\right)$
$\therefore 2\cos \frac{5x}{2}\cos \frac{x}{2}-2\sin x\cos \frac{x}{2}=0$
$\therefore 2\cos \left(x/2\right)\left[\cos \right(5x/2)-\sin x]=0$
$\cos \left(\frac{x}{2}\right)=0$ $\therefore \frac{x}{2}=(n+\frac{1}{2})\pi$
or $x=2n\pi +\pi$,
$n=0$, $x=\pi$ as $0\le x\le 2\pi$
$\cos \frac{5x}{2}=\sin x=\cos (\frac{\pi }{2}-x)$
$\frac{5x}{2}=2n\pi \pm (\frac{\pi }{2}-x)$
Taking $+$ive sign $\left(5x/2\right)+x=2n\pi +\pi /2$,
$\left(7x/2\right)=(4n+1)\left(\pi /2\right)$
$\therefore x=(4n+1)\pi /7$
$n=0,x=\pi /7,n=1,x=5\pi /7,n=2,x=9\pi /7$,
$n=3,x=13\pi /7$
Now taking $-$ive sign,
$\left(5x/2\right)-x=2n\pi -\pi /2$
or $\left(3x/2\right)=(4n-1)\pi /2$
$\therefore x=(4n-1)\pi /3$,
$\therefore$ For $n=1$, $x=\pi$.
$\therefore x=\frac{\pi }{7},\frac{5\pi }{7},\pi ,\frac{9\pi }{7},\frac{13\pi }{7}$.