The principal solutions of the equation sec x | vedclass

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Question

(A) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given equation: $\sec x + 	an x = 2 \cos x$$\Rightarrow \frac{1}{\cos x

Vedclass · Accepted Answer

$\frac{\pi}{6}, \frac{5 \pi}{6}$

Vedclass · Answer

(A) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given equation: $\sec x + 	an x = 2 \cos x$$\Rightarrow \frac{1}{\cos x} + \frac{\sin x}{\cos x} = 2 \cos x$$\Rightarrow 1 + \sin x = 2 \cos^2 x$$\Rightarrow 1 + \sin x = 2(1 - \sin^2 x)$$\Rightarrow 1 + \sin x = 2(1 - \sin x)(1 + \sin x)$Since $\sin x = -1$ makes $\cos x = 0$,which is undefined,we have $1 + \sin x 
eq 0$.Dividing both sides by $(1 + \sin x)$,we get:$1 = 2(1 - \sin x)$$\Rightarrow 1 = 2 - 2 \sin x$$\Rightarrow 2 \sin x = 1$$\Rightarrow \sin x = \frac{1}{2}$In the interval $[0, 2 \pi)$,the solutions are $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.

The principal solutions of the equation $\sec x + \tan x = 2 \cos x$ are

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