Question

Number of solutions of the trigonometric equation \[ 2 \tan 2\theta - \cot 2\theta + 1 = 0 \quad \text{lying in the interval} \quad [0, \pi] \]

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

When solving trigonometric equations, use algebraic manipulations and trigonometric identities to reduce the equation to a solvable form. Then, apply the relevant inverse trigonometric functions to find the solutions.

Answer 1

The Correct Option is C

We are given the equation: \[ 2 \tan 2\theta - \cot 2\theta + 1 = 0. \] We need to find the number of solutions in the interval \( [0, \pi] \). Step 1: Solve the equation Rearrange the equation: \[ 2 \tan 2\theta = \cot 2\theta - 1. \] Using the identity \( \cot x = \frac{1}{\tan x} \), we substitute \( \cot 2\theta = \frac{1}{\tan 2\theta} \) into the equation: \[ 2 \tan 2\theta = \frac{1}{\tan 2\theta} - 1. \] Multiply through by \( \tan 2\theta \): \[ 2 \tan^2 2\theta = 1 - \tan 2\theta. \] Rearrange the terms: \[ 2 \tan^2 2\theta + \tan 2\theta - 1 = 0. \] Let \( x = \tan 2\theta \), so we get the quadratic equation: \[ 2x^2 + x - 1 = 0. \] Step 2: Solve the quadratic equation Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 2, b = 1, c = -1 \). Substituting these values: \[ x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}. \] Thus, \( x = \frac{2}{4} = \frac{1}{2} \) or \( x = \frac{-4}{4} = -1 \). Step 3: Solve for \( \theta \) For \( \tan 2\theta = \frac{1}{2} \), solve \( 2\theta = \tan^{-1}\left( \frac{1}{2} \right) \). This gives two solutions in \( [0, \pi] \). For \( \tan 2\theta = -1 \), solve \( 2\theta = \tan^{-1}(-1) \), which also gives two solutions in \( [0, \pi] \). Thus, the total number of solutions is 4.