The numbers of solution (s) of the equation l | vedclass

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Question

$\left(\frac{2 \sin x-1}{2 \sin x}\right) \cos ^{2} 2 x =\frac{2 \sin ^{2} x-3 \sin x+1}{\sin x} $

$=\frac{(2 \sin x-1)(\sin x-1)}{\sin x} $

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Vedclass · Accepted Answer

$4$

Vedclass · Answer

$\left(\frac{2 \sin x-1}{2 \sin x}\right) \cos ^{2} 2 x =\frac{2 \sin ^{2} x-3 \sin x+1}{\sin x} $

$=\frac{(2 \sin x-1)(\sin x-1)}{\sin x} $

$ \Rightarrow \mathop {\mathop {\sin x = \frac{1}{2}}\limits_ \Downarrow  }\limits_{4\,solutions} {\rm{ }}\,\,\,\,\,{\rm{or }}\,\,\,\,\mathop {\mathop {\,\frac{1}{2}{{\cos }^2}2x}\limits_{ \ge 0}  = \mathop {\sin x - 1}\limits_{ \le 0} }\limits_{{\rm{Hence no solution }}} $

The numbers of solution $(s)$ of the equation $\left( {1 - \frac{1}{{2\,\sin x}}} \right){\cos ^2}\,2x\, = \,2\,\sin x\, - \,3\, + \,\frac{1}{{\sin x}}$ in $[0,4\pi ]$ is

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