If the sum of solutions of the system of equa | vedclass

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Question

$2 \sin ^{2} 	heta-\cos 2 	heta=0$

$2 \sin ^{2} 	heta-\left(1-2 \sin ^{2} 	hetaight)=0$

$\sin ^{2} 	heta=\left(\frac{1}{2}ight)^{2}$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$2 \sin ^{2} 	heta-\cos 2 	heta=0$

$2 \sin ^{2} 	heta-\left(1-2 \sin ^{2} 	hetaight)=0$

$\sin ^{2} 	heta=\left(\frac{1}{2}ight)^{2}$

$	heta=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$

$2 \cos ^{2} 	heta+3 \sin 	heta=0$

$2 \sin ^{2} 	heta-3 \sin 	heta-2=0$

$	herefore \sin 	heta=-\frac{1}{2}$

$	heta=\frac{7 \pi}{6}, \frac{11 \pi}{6}$

So, the common solution is

$	heta=\frac{7 \pi}{6}, \frac{11 \pi}{6}$

$	ext { Sum }=\frac{7 \pi+11 \pi}{6}=3 \pi= k \pi$

$K =3$

If the sum of solutions of the system of equations $2 \sin ^{2} \theta-\cos 2 \theta=0$ and $2 \cos ^{2} \theta+3 \sin \theta=0$ in the interval $[0,2 \pi]$ is $k \pi$, then $k$ is equal to.

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