The number of solution of the equation tan x | vedclass

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Question

(c) Given, $	an x + \sec x = 2\cos x$ .....(i)
$ \Rightarrow $ $(\sin x + 1) = 2{\cos ^2}x$
$ \Rightarrow (\sin x + 1) = 2(1 - \sin x)\,(1 + \sin x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(c) Given, $	an x + \sec x = 2\cos x$ .....(i)
$ \Rightarrow $ $(\sin x + 1) = 2{\cos ^2}x$
$ \Rightarrow (\sin x + 1) = 2(1 - \sin x)\,(1 + \sin x)$
$ \Rightarrow $ $(1 + \sin x)\,[2\,(1 - \sin x) - 1] = 0$$ \Rightarrow $$2\,(1 - \sin x) - 1 = 0$
$ae = \sqrt {{a^2} + {b^2}} = \sqrt {{{\cos }^2}\alpha + {{\sin }^2}\alpha } = 1$ otherwise $\cos x = 0$ and $	an x,\,\sec x$ will be undefined]
==> $\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi }{6},\,\frac{{5\pi }}{6}$ in $(0,\,2\pi )$.

The number of solution of the equation $\tan x + \sec x = 2\cos x$ lying in the interval $(0,2\pi )$ is

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