The number of solutions of the equation 2 sin | vedclass

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(B) Given equation: $2 \sin^2 	heta - \sin 	heta \cos 	heta - 3 \cos^2 	heta = 0$. Dividing by $\cos^2 	heta$ (assuming $\cos 	heta 
eq 0$): $2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given equation: $2 \sin^2 	heta - \sin 	heta \cos 	heta - 3 \cos^2 	heta = 0$. Dividing by $\cos^2 	heta$ (assuming $\cos 	heta 
eq 0$): $2 	an^2 	heta - 	an 	heta - 3 = 0$. Let $x = 	an 	heta$,then $2x^2 - x - 3 = 0$. $(2x - 3)(x + 1) = 0$. So,$	an 	heta = \frac{3}{2}$ or $	an 	heta = -1$. For $	an 	heta = -1$,$	heta = -\frac{\pi}{4}$ and $	heta = \frac{3\pi}{4}$ in $(-\pi, \pi)$. For $	an 	heta = \frac{3}{2}$,$	heta = \arctan(\frac{3}{2})$ and $	heta = \arctan(\frac{3}{2}) - \pi$ in $(-\pi, \pi)$. Total number of solutions is $4$.

The number of solutions of the equation $2 \sin^2 \theta - 3 \cos^2 \theta = \sin \theta \cos \theta$ lying in the interval $(-\pi, \pi)$ is

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