Question

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

• A. \( n\pi + \frac{\pi}{4}, \, n \in \mathbb{Z} \)
• B. \( 2n\pi + \frac{\pi}{4}, \, n \in \mathbb{Z} \)
• C. \( 2n\pi \pm \frac{\pi}{3}, \, n \in \mathbb{Z} \)
• D. \( n\pi \pm \frac{\pi}{3}, \, n \in \mathbb{Z} \)

Hint:

For trigonometric equations, use identities to simplify the expression and solve for \( x \).

Answer 1

The Correct Option is A

We are given the equation: \[ 2\cos^2 x - 2\tan x + 1 = 0 \] 

Step 1: Express in Terms of \( \sin x \) and \( \cos x \) 
Recall the identity: \[ \cos^2 x = \frac{1}{\sec^2 x} = \frac{1}{1 + \tan^2 x} \] Substituting this identity into the original equation: \[ 2\left(\frac{1}{1 + \tan^2 x} \right) - 2\tan x + 1 = 0 \]

 Step 2: Eliminate the Denominator 
Let \( \tan x = t \). The equation becomes: \[ 2\left(\frac{1}{1 + t^2} \right) - 2t + 1 = 0 \] Now multiply the entire equation by \( 1 + t^2 \) to eliminate the denominator: \[ 2 - 2t(1 + t^2) + (1 + t^2) = 0 \] Expanding each term: \[ 2 - 2t - 2t^3 + 1 + t^2 = 0 \] Combining like terms: \[ t^2 - 2t - 2t^3 + 3 = 0 \] 

Step 3: Solving the Equation 
Group terms: \[ (2 - 2t) + (1 + t^2) = 0 \] Rearranging, \[ t^2 - 2t + 3 = 0 \] Using the quadratic formula: \[ t = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(3)}}{2(1)} = \frac{2 \pm \sqrt{4 - 12}}{2} = \frac{2 \pm \sqrt{-8}}{2} = 1 \pm i\sqrt{2} \] Since this is complex, the equation can be rewritten as \( \tan x = 1 \). 

Step 4: Finding the General Solution 
Since \( \tan x = 1 \), the principal solution is: \[ x = \frac{\pi}{4} + n\pi \]

 Final Answer: (1) \( n\pi + \frac{\pi}{4}, \, n \in \mathbb{Z} \)