The general solution of 2 cos^2 x - 2 tan x + | vedclass

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Question

(A) Given equation: $2 \cos^2 x - 2 	an x + 1 = 0$. We know that $2 \cos^2 x = 1 + \cos 2x$. Substituting this,we get $1 + \cos 2x - 2 	an x + 1 = 0

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$n \pi + \frac{\pi}{4}, n \in \mathbb{Z}$

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(A) Given equation: $2 \cos^2 x - 2 	an x + 1 = 0$. We know that $2 \cos^2 x = 1 + \cos 2x$. Substituting this,we get $1 + \cos 2x - 2 	an x + 1 = 0$,which simplifies to $\cos 2x - 2 	an x + 2 = 0$. Using $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we have $\frac{1 - 	an^2 x}{1 + 	an^2 x} - 2(	an x - 1) = 0$. Factoring out $(	an x - 1)$,we get $\frac{-(1 - 	an x)(1 + 	an x)}{1 + 	an^2 x} - 2(	an x - 1) = 0$. $(	an x - 1) [\frac{1 + 	an x}{1 + 	an^2 x} + 2] = 0$. Case $1$: $	an x - 1 = 0$ $\Rightarrow 	an x = 1$ $\Rightarrow x = n \pi + \frac{\pi}{4}$. Case $2$: $\frac{1 + 	an x + 2 + 2 	an^2 x}{1 + 	an^2 x} = 0 \Rightarrow 2 	an^2 x + 	an x + 3 = 0$. The discriminant $D = 1^2 - 4(2)(3) = 1 - 24 = -23 Thus,the general solution is $x = n \pi + \frac{\pi}{4}, n \in \mathbb{Z}$.

The general solution of $2 \cos^2 x - 2 \tan x + 1 = 0$ is

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