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

• A

  $2$
• Correct answer
  B

  $3$
• C

  $4$
• D

  $5$

Solution

$\left(\text{sin x}+\text{sin 5x}\right)+\text{sin 3x}=0$
$\text{2 sin 3x cos 2x}+\text{sin 3x}=0$
$\text{sin 3x}\left(\text{2 cos 2x}+1\right)=0$
$\text{sin 3x}=0$ and $\text{2 cos 2x}+1=0$
$\sin 3x=0$
Graph f(x) = sin 3x

$x=\frac{2\pi }{3}$ or $x=\pi$ or $x=\frac{4\pi }{3}\text{,}$
$\therefore \text{sin}3x=0$ have 3 solutions
Graph f(x) = cos 2x

$\text{cos}2x=-\frac{1}{2}$
$\therefore x=\frac{2\pi }{3} and\frac{4\pi }{3}$
So the solution of given equation in $[\frac{\pi }{2},\frac{3\pi }{2}]$ are $\frac{2\pi }{3} and \frac{4\pi }{3}$

Hence, the correct answer is option: 'B'.