Q

# If sinθ – sin2θ + sin3θ = 0. Then number of solutions in

$If\:\sin \left(\theta \right)-\sin \left(2\theta \right)+\sin \left(3\theta \right)=0.\:Then\:number\:of\:solutions\:in\:[0,\frac{\pi}{2})$

Views

$\:\sin \left(\theta \right)-\sin \left(2\theta \right)+\sin \left(3\theta \right)=0.$

Now , As we know,

$\sin A +\sin B=2\sin\left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$

So,

$\sin \theta +\sin 3\theta-\sin 2\theta=2\sin\left ( \frac{\theta +3\theta}{2} \right )\cos\left ( \frac{\theta-3\theta}{2} \right )-\sin 2\theta$

$\sin \theta +\sin 3\theta-\sin 2\theta=2\sin 2\theta\cos\theta-\sin 2\theta$

So the equations become,

$2\sin 2\theta\cos\theta-\sin 2\theta=0$

$2\sin 2\theta\cos\theta=\sin 2\theta$

$2\cos \theta = 1$  provided  $\sin 2\theta \neq0$

$\cos\theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$.

Hence the number of solutions in  $[0,\pi/2]$ is 1.

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