The number of solutions of the equation left( | vedclass

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Question

(C) The given equation is $\left( \frac{2 \sin x - 1}{2 \sin x} \right) \cos^2 2x = \frac{2 \sin^2 x - 3 \sin x + 1}{\sin x}$.Factoring the numerator

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The given equation is $\left( \frac{2 \sin x - 1}{2 \sin x} ight) \cos^2 2x = \frac{2 \sin^2 x - 3 \sin x + 1}{\sin x}$.Factoring the numerator on the right side: $2 \sin^2 x - 3 \sin x + 1 = (2 \sin x - 1)(\sin x - 1)$.So,$\left( \frac{2 \sin x - 1}{2 \sin x} ight) \cos^2 2x = \frac{(2 \sin x - 1)(\sin x - 1)}{\sin x}$.This implies either $2 \sin x - 1 = 0$ or $\frac{1}{2} \cos^2 2x = \sin x - 1$.Case $1$: $\sin x = \frac{1}{2}$. In $[0, 4\pi]$,$\sin x = \frac{1}{2}$ has $4$ solutions $(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6})$.Case $2$: $\frac{1}{2} \cos^2 2x = \sin x - 1$. Since $\frac{1}{2} \cos^2 2x \ge 0$ and $\sin x - 1 \le 0$,the only possibility is $\cos^2 2x = 0$ and $\sin x - 1 = 0$.If $\sin x = 1$,then $\cos 2x = 1 - 2 \sin^2 x = 1 - 2(1) = -1$,so $\cos^2 2x = 1 
eq 0$. Thus,no solution exists for Case $2$.Therefore,the total number of solutions is $4$.

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

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