Discussion Forum

The number  of solution of $\sin x$ +sin 3x+sin 5x=0  in the interval  $\left[ \frac{\pi}{2},3\frac{\pi}{2}\right]$  is 

Answer:

Explanation:

 We have,

 $\sin x$ +sin 3x+sin 5x=0

$\Rightarrow$     $\sin x+\sin 5x+\sin x$=0

$\Rightarrow$  2$\sin 3x \cos 2x+\sin 3x$=0

$\Rightarrow$    $\sin 3x (2\cos 2x +1)$=0

$\Rightarrow$   $\sin 3x=0$  or $\cos 2x=-\frac{1}{2}=\cos \frac{2 \pi}{3}$

$\Rightarrow$ $3x=n\pi$  or   $2n=2n\pi\pm\frac{2\pi}{3}$

$\Rightarrow$    $x=\frac{n\pi}{3}$   or   $x=n\pi\pm\frac{\pi}{3}$  

  But, it is given that

$x \epsilon \left[\frac{\pi}{2},\frac{3\pi}{2}\right]$

$\therefore$      $x=\frac{2\pi}{3},\pi,\frac{4\pi}{3}$

$\therefore$    Number of solutions is 3.