Q64 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

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

  • Correct answer
    A
    $n \pi+\frac{\pi}{4}, n \in \mathrm{Z}$
  • B
    $2 n \pi \pm \frac{\pi}{4}, n \in \mathrm{Z}$
  • C
    $2 n \pi \pm \frac{\pi}{3}, n \in \mathrm{Z}$
  • D
    $n \pi \pm \frac{\pi}{3}, n \in \mathrm{Z}$

Solution

Given, $2 \cos ^2 x-2 \tan x+1=0$ $\begin{aligned} & \Rightarrow 2 \cos ^2 x-1-2 \tan x+2=0 \\ & \Rightarrow \cos 2 x=2(\tan x-1) \Rightarrow \frac{1-\tan ^2 x}{1+\tan ^2 x}=2(\tan x-1)\end{aligned}$ $\Rightarrow 1+\tan x=-2\left(1+\tan ^2 x\right) \quad(\because$ if $\tan x-1 \neq 0)$ $\Rightarrow 2 \tan ^2 x+\tan x+3=0$ Since, $1-4 \times 2 \times 3 \lt 0$, So, $x$ does not exists If $\tan x-1=0 \Rightarrow x=\frac{\pi}{4}+n \pi$.

Hence, the correct answer is option: 'A'.

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