If the equation 2 tan x sin x - 2 tan x + cos | vedclass

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(B) Given equation: $2 	an x \sin x - 2 	an x + \cos x = 0$$\frac{2 \sin^2 x}{\cos x} - \frac{2 \sin x}{\cos x} + \cos x = 0$$\frac{2 \sin^2 x - 2 \

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$1$

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(B) Given equation: $2 	an x \sin x - 2 	an x + \cos x = 0$$\frac{2 \sin^2 x}{\cos x} - \frac{2 \sin x}{\cos x} + \cos x = 0$$\frac{2 \sin^2 x - 2 \sin x + \cos^2 x}{\cos x} = 0$Since $\cos^2 x = 1 - \sin^2 x$,we have:$\frac{2 \sin^2 x - 2 \sin x + 1 - \sin^2 x}{\cos x} = 0$$\frac{\sin^2 x - 2 \sin x + 1}{\cos x} = 0$$\frac{(\sin x - 1)^2}{\cos x} = 0$This implies $\sin x = 1$ and $\cos x 
eq 0$.For $\sin x = 1$,$x = \frac{\pi}{2} + 2n\pi$ where $n \in \mathbb{Z}$.In the interval $[0, k\pi]$,the solutions are $x = \frac{\pi}{2}, \frac{5\pi}{2}, \dots$If $k=1$,interval is $[0, \pi]$,solution is $x = \frac{\pi}{2}$ ($1$ solution). So $k=1$ is a solution.If $k=2$,interval is $[0, 2\pi]$,solution is $x = \frac{\pi}{2}$ ($1$ solution). Here $k=2$ but number of solutions is $1 
eq k$. So $k=2$ is not a solution.Thus,only $k=1$ satisfies the condition.

If the equation $2 \tan x \sin x - 2 \tan x + \cos x = 0$ has $k$ solutions in the interval $[0, k\pi]$,then the number of integral values of $k$ is-

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