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Q.

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

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a

$(2 n + 1) \frac{π}{3}, n \in Z$

b

$(n + 1) \frac{π}{3}, n \in Z$

c

$n π + \frac{π}{3}, n \in Z$

d

$n π + \frac{π}{4}, n \in Z$

answer is D.

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Detailed Solution

$\frac{1 - \tan^{2} x}{1 + \tan^{2} x} - 2 \tan x + 2 = 0$
$\begin{array}{l} \Rightarrow (1 - \tan^{2} x) - 2 (\tan x - 1) (1 + \tan^{2} x) = 0 \\ \Rightarrow (1 - \tan x) (2 \tan^{2} x + \tan x + 3) = 0 \\ \Rightarrow \tan x = 1 \Rightarrow \tan x = \tan \frac{π}{4} \\ \Rightarrow x = n π + \frac{π}{4} \end{array}$

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