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.

Answer 2

Find the number of solutions of the equation: \[ 2 \tan 2\theta - \cot 2\theta + 1 = 0 \quad \text{for} \quad \theta \in [0, \pi]. \]

Step 1: Express all terms in terms of \(\tan 2\theta\)
Recall that \(\cot 2\theta = \frac{1}{\tan 2\theta}\). Rewrite the equation as: \[ 2 \tan 2\theta - \frac{1}{\tan 2\theta} + 1 = 0. \]
Multiply through by \(\tan 2\theta\) (assuming \(\tan 2\theta \neq 0\)): \[ 2 (\tan 2\theta)^2 - 1 + \tan 2\theta = 0. \]

Step 2: Rearrange into quadratic form
Let \( t = \tan 2\theta \). Then, the equation becomes: \[ 2 t^2 + t - 1 = 0. \]

Step 3: Solve the quadratic equation
Using the quadratic formula, \[ t = \frac{-1 \pm \sqrt{1 + 8}}{2 \times 2} = \frac{-1 \pm 3}{4}. \] So, \[ t_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}, \quad t_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1. \]

Step 4: Find solutions for \(\theta\)
Since \( t = \tan 2\theta \), \[ \tan 2\theta = \frac{1}{2} \quad \text{or} \quad \tan 2\theta = -1. \] For \(\theta \in [0, \pi]\), the range for \(2\theta\) is \([0, 2\pi]\). - For \(\tan 2\theta = \frac{1}{2}\), general solutions are: \[ 2\theta = \tan^{-1}\left(\frac{1}{2}\right) + n\pi, \quad n = 0, 1. \] Thus, \[ \theta = \frac{1}{2} \tan^{-1}\left(\frac{1}{2}\right), \quad \text{and} \quad \theta = \frac{1}{2}\left(\tan^{-1}\left(\frac{1}{2}\right) + \pi\right). \] - For \(\tan 2\theta = -1\), general solutions are: \[ 2\theta = \tan^{-1}(-1) + n\pi = -\frac{\pi}{4} + n\pi, \quad n = 1, 2. \] Taking only values in \([0, 2\pi]\), \[ 2\theta = \frac{3\pi}{4}, \quad \text{and} \quad 2\theta = \frac{7\pi}{4}. \] Thus, \[ \theta = \frac{3\pi}{8}, \quad \text{and} \quad \theta = \frac{7\pi}{8}. \]

Step 5: Counting the number of valid solutions
Total solutions in \([0, \pi]\) are: - Two from \(\tan 2\theta = \frac{1}{2}\)
- Two from \(\tan 2\theta = -1\)
Hence, the total number of solutions is \[ \boxed{4}. \]