The number of solutions of the equation tan x | vedclass

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(C) Given equation: $	an x + \sec x = 2 \cos x$Using $	an x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$,we have:$\frac{\sin x + 1}{\cos

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$2$

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(C) Given equation: $	an x + \sec x = 2 \cos x$Using $	an x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$,we have:$\frac{\sin x + 1}{\cos x} = 2 \cos x$$\sin x + 1 = 2 \cos^2 x$Since $\cos^2 x = 1 - \sin^2 x$,the equation becomes:$\sin x + 1 = 2(1 - \sin^2 x)$$\sin x + 1 = 2 - 2 \sin^2 x$$2 \sin^2 x + \sin x - 1 = 0$Factoring the quadratic equation:$2 \sin^2 x + 2 \sin x - \sin x - 1 = 0$$2 \sin x(\sin x + 1) - 1(\sin x + 1) = 0$$(2 \sin x - 1)(\sin x + 1) = 0$This gives $\sin x = \frac{1}{2}$ or $\sin x = -1$.For $x \in [0, \pi]$,$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.If $\sin x = -1$,then $x = \frac{3\pi}{2}$,which is outside the interval $[0, \pi]$.Also,at $x = \frac{\pi}{2}$,$	an x$ and $\sec x$ are undefined,so $x = \frac{\pi}{2}$ is not a solution.Thus,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.The number of solutions is $2$.

The number of solutions of the equation $\tan x + \sec x = 2 \cos x$ for $x \in [0, \pi]$ is

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