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

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(D) Given equation: $\cos 2x - 2 	an x + 2 = 0$Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$\frac{1 - 	an^2 x}{1 + 	an^

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

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(D) Given equation: $\cos 2x - 2 	an x + 2 = 0$Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$\frac{1 - 	an^2 x}{1 + 	an^2 x} - 2 	an x + 2 = 0$Multiplying by $(1 + 	an^2 x)$:$(1 - 	an^2 x) - 2 	an x(1 + 	an^2 x) + 2(1 + 	an^2 x) = 0$$1 - 	an^2 x - 2 	an x - 2 	an^3 x + 2 + 2 	an^2 x = 0$$-2 	an^3 x + 	an^2 x - 2 	an x + 3 = 0$$2 	an^3 x - 	an^2 x + 2 	an x - 3 = 0$Let $	an x = t$,then $2t^3 - t^2 + 2t - 3 = 0$.By inspection,$t = 1$ is a root: $2(1)^3 - (1)^2 + 2(1) - 3 = 2 - 1 + 2 - 3 = 0$.Dividing by $(t - 1)$,we get $(t - 1)(2t^2 + t + 3) = 0$.The quadratic $2t^2 + t + 3$ has discriminant $D = 1^2 - 4(2)(3) = 1 - 24 = -23 Thus,$	an x = 1 = 	an \frac{\pi}{4}$.The general solution is $x = n\pi + \frac{\pi}{4}, n \in Z$.

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

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