The number of solutions of the equation tan x | vedclass

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(C) Given equation: $	an x + \sec x = 2\cos x$ $\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$ $\Rightarrow \frac{\sin x + 1}{\cos x

Vedclass · Accepted Answer

$2$

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(C) Given equation: $	an x + \sec x = 2\cos x$ $\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$ $\Rightarrow \frac{\sin x + 1}{\cos x} = 2\cos x$ $\Rightarrow \sin x + 1 = 2\cos^2 x$ $\Rightarrow \sin x + 1 = 2(1 - \sin^2 x)$ $\Rightarrow \sin x + 1 = 2(1 - \sin x)(1 + \sin x)$ $\Rightarrow (1 + \sin x) [1 - 2(1 - \sin x)] = 0$ $\Rightarrow (1 + \sin x) (2\sin x - 1) = 0$ Case $1$: $1 + \sin x = 0 \Rightarrow \sin x = -1$. At $\sin x = -1$,$\cos x = 0$,so $	an x$ and $\sec x$ are undefined. Thus,this is not a solution. Case $2$: $2\sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2}$. In the interval $(0, 2\pi)$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. Both values are valid. Therefore,there are $2$ 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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