Question

The number of solutions of the equation
$\sin \left(9x\right)+\sin \left(3x\right)=0$
in the closed interval $[0,2\pi ]$ is

• A. $7$
• B. $13$
• C. $19$
• D. $25$

Answer 1

The correct option is B $13$

$\sin \left(9x\right)+\sin \left(3x\right)=0\Rightarrow 2\sin 6x\cos 3x=0$
So, $\sin 6x=0$ or $\cos 3x=0$
$\sin 6x=0\Rightarrow 6x=0,\pi ,2\pi ,...........,12\pi \Rightarrow x=0,\frac{\pi }{6},\frac{2\pi }{6},...........\frac{12\pi }{6}$
From this we will get $13$ values
$\cos 3x=0\Rightarrow 3x=\frac{\pi }{2},\frac{3\pi }{2},........,\frac{11\pi }{2}\Rightarrow x=\frac{\pi }{6},\frac{3\pi }{6},........,\frac{11\pi }{6}$
This all values are already covered in above solutions
Hence, the number of solutions is $13$