The number of solutions of tan x + sec x = 2 | vedclass

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Question

(D) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given: $	an x + \sec x = 2 \cos x$$\Rightarrow \frac{\sin x}{\cos x} +

Vedclass · Accepted Answer

$2$

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(D) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given: $	an x + \sec x = 2 \cos x$$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$$\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)$$\Rightarrow (1 + \sin x)[2(1 - \sin x) - 1] = 0$Since $\sin x = -1$ implies $\cos x = 0$,which makes $	an x$ and $\sec x$ undefined,we have $1 + \sin x 
eq 0$.Therefore,$2(1 - \sin x) - 1 = 0$$\Rightarrow 2 - 2 \sin x - 1 = 0$$\Rightarrow 2 \sin x = 1$$\Rightarrow \sin x = \frac{1}{2}$In the interval $[0, 2 \pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$.Thus,the number of solutions is $2$.

The number of solutions of $\tan x + \sec x = 2 \cos x$ in $[0, 2 \pi]$ are

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