Let $\tan ^{-1}(\mathrm{x}) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, for $\mathrm{x} \in \mathbb{R}$. Then the number of real solutions of the equation $\sqrt{1+\cos (2 x)}=\sqrt{2} \tan ^{-1}(\tan x)$ in the set $\left(-\frac{3 \pi}{2},-\frac{\pi}{2}\right) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ is equal to

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Inverse Trigonometric Functions

Let

tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})

, for

x \in R

. Then the number of real solutions of the equation

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

in the set

(- \frac{3 π}{2}, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2})

is equal to

Answer: 3

The equation is

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

.

Using

1 + cos (2 x) = 2 cos^{2} (x)

, we get:

2 ∣ cos x ∣ = 2 tan^{- 1} (tan x)

∣ cos x ∣ = tan^{- 1} (tan x)

Note that

tan^{- 1} (tan x) = x

only when

x \in (- \frac{π}{2}, \frac{π}{2})

.

For

x \in (- \frac{π}{2}, \frac{π}{2})

∣ cos x ∣ = x

has one solution near

x \approx 0.739

.

For

x \in (\frac{π}{2}, \frac{3 π}{2})

tan^{- 1} (tan x) = x - π

, so

∣ cos x ∣ = x - π

. This gives one solution.

For

x \in (- \frac{3 π}{2}, - \frac{π}{2})

tan^{- 1} (tan x) = x + π

, so

∣ cos x ∣ = x + π

. This gives one solution.

Total number of solutions = 3

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer