Question

Number of solutions of the equation \( \tan^2 x + 3\cot^2 x = 2\sec^2 x \) lying in the interval \( [0, 2\pi] \) is

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

When solving trigonometric equations, a good first step is to use identities to express the equation in terms of a single trigonometric function. Substituting a variable (like \(y = \tan^2 x\)) can make the underlying algebraic structure (often a quadratic) easier to see and solve.

Answer 1

The Correct Option is B

We can express the entire equation in terms of \( \tan x \).
We use the identities \( \cot^2 x = \frac{1}{\tan^2 x} \) and \( \sec^2 x = 1 + \tan^2 x \).
The equation becomes \( \tan^2 x + \frac{3}{\tan^2 x} = 2(1 + \tan^2 x) \).
Let \( y = \tan^2 x \). The equation in terms of y is \( y + \frac{3}{y} = 2(1+y) \).
Note that \( y \ge 0 \) since it is a square. Also, \( y \ne 0 \) for \( \cot x \) to be defined.
Multiply the equation by \(y\) to clear the fraction:
\( y^2 + 3 = 2y(1+y) = 2y + 2y^2 \).
Rearrange the terms to form a quadratic equation:
\( y^2 + 2y - 3 = 0 \).
Factor the quadratic equation: \( (y+3)(y-1) = 0 \).
This gives two possible solutions for y: \( y = -3 \) or \( y = 1 \).
Since \( y = \tan^2 x \), it cannot be negative. So we must discard \( y = -3 \).
The only valid solution is \( \tan^2 x = 1 \).
This leads to two possibilities: \( \tan x = 1 \) or \( \tan x = -1 \).
In the interval \( [0, 2\pi] \):
For \( \tan x = 1 \), the solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
For \( \tan x = -1 \), the solutions are \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \).
In total, there are four solutions.