The number of solutions of the equation tan x | vedclass

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(B) Given equation: $	an x + \sec x = 2 \cos x$We can rewrite this as: $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$Multiplying both sides by

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$2$

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(B) Given equation: $	an x + \sec x = 2 \cos x$We can rewrite this as: $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$Multiplying both sides by $\cos x$ (where $\cos x 
eq 0$): $\sin x + 1 = 2 \cos^2 x$Using the identity $\cos^2 x = 1 - \sin^2 x$: $\sin x + 1 = 2(1 - \sin^2 x)$Rearranging the terms: $2 \sin^2 x + \sin x - 1 = 0$Factoring the quadratic equation: $(2 \sin x - 1)(\sin x + 1) = 0$This gives two possible values: $\sin x = \frac{1}{2}$ or $\sin x = -1$Case $1$: If $\sin x = -1$,then $x = \frac{3 \pi}{2}$. At $x = \frac{3 \pi}{2}$,$\cos x = 0$,which makes $	an x$ and $\sec x$ undefined. Thus,this is not a valid solution.Case $2$: If $\sin x = \frac{1}{2}$,then $x = \frac{\pi}{6}$ or $x = \frac{5 \pi}{6}$. Both values are within the interval $[0, 2 \pi]$ and $\cos x 
eq 0$ at these points.Therefore,there are $2$ valid solutions.

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

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