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  • 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})

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\:\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.

 

Posted by

Pankaj Sanodiya

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