Question

Let tan^-1 x ∈(\(-\frac π2\) \(\frac π2\))  for x ∈ R Then the number of real solutions of the equation √1 + cos (2x) = √2 tan -1 (tan x) in the set (- 3π/2, - π/2) ∪ (- π/2, π/2) ∪ (π/2, 3π/2) is equal to 

Answer 1

Correct Answer: 3

Real Solutions of the Given Equation 

Given: The equation is:

\(\sqrt{1 + \cos(2x)} = \sqrt{2} \tan^{-1}(\tan x)\)

We are tasked with finding the number of real solutions of this equation in the set \( \left( -\frac{3\pi}{2}, -\frac{\pi}{2} \right) \cup \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \cup \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \).

Solution:

First, recall that the domain of \( \tan^{-1} x \) is \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), so we must consider this constraint for the equation \( \tan^{-1} (\tan x) \).

Next, consider the behavior of both sides of the equation:

• For \( \sqrt{1 + \cos(2x)} \), we know that this is a periodic function with a period of \( \pi \), as the cosine function repeats every \( \pi \). Therefore, we expect solutions within the given intervals.
• For \( \sqrt{2} \tan^{-1} (\tan x) \), we note that \( \tan^{-1} (\tan x) = x \) for \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), but it behaves differently for values outside this range due to periodicity.

By examining the equation over the intervals \( \left( -\frac{3\pi}{2}, -\frac{\pi}{2} \right) \), \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), and \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \), we can determine the number of solutions.

The number of real solutions of the equation in the given set is 3.

Answer 2

The correct answer is 3

Number of solutions = Number of intersection points = 3.