The number of solutions of the equation sin t | vedclass

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Question

(b)

We have, $\sin 	heta+\cos 	heta=\sin 2 	heta$ On squaring both sides, we get $\sin ^2 	heta+\cos ^2 	heta+\sin 2 	heta=\sin ^2 2 	heta$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b)

We have, $\sin 	heta+\cos 	heta=\sin 2 	heta$ On squaring both sides, we get $\sin ^2 	heta+\cos ^2 	heta+\sin 2 	heta=\sin ^2 2 	heta$ $\Rightarrow \sin ^2 2 	heta-\sin 2 	heta-1=0$

$\Rightarrow \quad \sin 2 	heta=\frac{1 \pm \sqrt{5}}{2}$

$\Rightarrow \sin 2 	heta=\frac{1-\sqrt{5}}{2} \Rightarrow \sin 2 	heta 
eq \frac{\sqrt{5}+1}{2}$

$	herefore$ Two solutions exist interval $[-\pi, \pi]$.

The number of solutions of the equation $\sin \theta+\cos \theta=\sin 2 \theta$ in the interval $[-\pi, \pi]$ is

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