જો 2{sin ^2}theta = 3cos theta , કે જ્યાં 0 l | vedclass

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Question

(b) $2 - 2{\cos ^2}	heta = 3\cos 	heta $

==> $2{\cos ^2} + 3\cos 	heta - 2 = 0$

==> $\cos 	heta = \frac{{ - 3 \pm \sqrt {9 + 16} }}{4}

Vedclass · Accepted Answer

$\frac{\pi }{3},\frac{{5\pi }}{3}$

Vedclass · Answer

(b) $2 - 2{\cos ^2}	heta = 3\cos 	heta $

==> $2{\cos ^2} + 3\cos 	heta - 2 = 0$

==> $\cos 	heta = \frac{{ - 3 \pm \sqrt {9 + 16} }}{4} = \frac{{ - 3 \pm 5}}{4}$

Neglecting $(-)$ sign, we get

$\cos 	heta = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} ight)$

$ \Rightarrow $ $	heta = 2n\pi \pm \frac{\pi }{3}$.

The values of $	heta $ between $0$ and $2\pi $ are $\frac{\pi }{3},{m{ }}\frac{{{m{5}}\pi }}{{m{3}}}$.

જો $2{\sin ^2}\theta = 3\cos \theta ,$ કે જ્યાં $0 \le \theta \le 2\pi $, તો $\theta = $

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