Let tan⁻¹(x) ∈ (-π/2, π/2), for x ∈ ℝ. Then the number of real solutions of the equation √(1 + cos(2

Q: Wait, why do we reject the negative root here? Or rather, why is $\sqrt{\cos^2 x} = |\cos x|$?

This is the most common pitfall in JEE trigonometry! Remember, the square root function $\sqrt{y}$ always returns a non-negative value by definition. If you write $\sqrt{\cos^2 x} = \cos x$, you are assuming $\cos x$ is always positive, which is false. The modulus $|\cos x|$ ensures that the output remains non-negative, preserving the mathematical truth of the equation.

Q: Why can't we just say $\tan^{-1}(\tan x) = x$ for all $x$?

That is a seductive trap! The function $\tan^{-1}(\tan x)$ is periodic with period $\pi$. It is only equal to $x$ when $x$ is in the principal branch $(-\frac{\pi}{2}, \frac{\pi}{2})$. Outside this, the graph shifts vertically by multiples of $\pi$. Always check the interval before simplifying inverse trigonometric functions.

Q: How do I know how many times the graphs intersect without solving for $x$?

Great question! You don't need the exact value of $x$ to count the solutions. By sketching the graphs, you can see the 'bumps' of $|\cos x|$ and the 'lines' of $\tan^{-1}(\tan x)$. Since the slope of the line is 1 and the curvature of the cosine bump is bounded, you can visually confirm that each line segment intersects each bump exactly once. Geometry is often faster and safer than algebra!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks for the number of real solutions to the equation:

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

To solve this, we must first simplify the trigonometric expression on the left-hand side.

### Taming the Left Hand Side

Recall the trigonometric identity $1 + \cos(2x) = 2\cos^2 x$. Substituting this into the equation yields:

$$\sqrt{2\cos^2 x} = \sqrt{2} 	an^{-1}(	an x)$$

Crucially, we must remember that $\sqrt{a^2} = |a|$. Therefore, the left-hand side (LHS) simplifies to:

$$\sqrt{2}|\cos x| = \sqrt{2} 	an^{-1}(	an x)$$

Dividing both sides by $\sqrt{2}$, we obtain the simplified master equation:

$$|\cos x| = 	an^{-1}(	an x)$$

### Unmasking the Right Hand Side

The function $g(x) = 	an^{-1}(	an x)$ is a periodic sawtooth wave. It is defined as $g(x) = x - n\pi$ for $x \in ((n - \frac{1}{2})\pi, (n + \frac{1}{2})\pi)$, where $n$ is an integer.

We analyze the intersections of $f(x) = |\cos x|$ and $g(x)$ across different intervals:

### The Graphical Symphony

**1. The Central Interval $(-\frac{\pi}{2}, \frac{\pi}{2})$:**
In this range, $g(x) = x$. We solve $|\cos x| = x$. At $x=0$, $|\cos 0| = 1$ and $x=0$. At $x=\frac{\pi}{2}$, $|\cos x| = 0$ and $x=\frac{\pi}{2}$. Since $|\cos x|$ is strictly decreasing and $x$ is strictly increasing on this interval, there is exactly **one intersection**.

**2. The Right Interval $(\frac{\pi}{2}, \frac{3\pi}{2})$:**
In this range, $g(x) = x - \pi$. We solve $|\cos x| = x - \pi$. At $x=\pi$, $|\cos \pi| = 1$ and $x - \pi = 0$. At $x=\frac{3\pi}{2}$, $|\cos x| = 0$ and $x - \pi = \frac{\pi}{2}$. Because the functions cross each other's values, there is exactly **one intersection**.

**3. The Left Interval $(-\frac{3\pi}{2}, -\frac{\pi}{2})$:**
In this range, $g(x) = x + \pi$. We solve $|\cos x| = x + \pi$. By applying the same logic as the previous interval, we find exactly **one intersection** point here as well.

### Conclusion: The Beauty of Symmetry

By partitioning the domain into these three intervals, we have identified one solution in each.

Summing these up, we find that there are $1 + 1 + 1 = 3$ real solutions. This problem serves as a reminder that in JEE Advanced, the most powerful tool is the ability to visualize the behavior of functions.

**Total number of real solutions = 3**

Enter Numerical Value:

Visualized Solution

The Equation

Simplifying LHS

The Modulus Trap

Simplified Equation

Graphical Approach Setup

Graphing $f (x)$

Analyzing $g (x)$ (Central Interval)

Intersection in Central Interval

Analyzing $g (x)$ (Right Interval)

Intersection in Right Interval

Analyzing $g (x)$ (Left Interval)

Intersection in Left Interval

Final Count

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Taming the Left Hand Side

Unmasking the Right Hand Side

The Graphical Symphony

Conclusion

Similar Questions

For $x \in (- 1, 1]$ , the number of solutions of the equation $sin^{- 1} x = 2 tan^{- 1} x$ is equal to _________.

If the sum of all the solutions of $tan^{- 1} (\frac{2 x}{1 - x ^{2}}) + cot^{- 1} (\frac{1 - x ^{2}}{2 x}) = \frac{π}{3}$ , $- 1 < x < 1, x \neq = 0$ , is $α - \frac{4}{3}$ , then $α$ is equal to

For any $y \in R$ , let $cot^{- 1} (y) \in (0, π)$ and $tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2})$ . Then the sum of all the solutions of the equation $tan^{- 1} (\frac{6 y}{9 - y ^{2}}) + cot^{- 1} (\frac{9 - y ^{2}}{6 y}) = \frac{2 π}{3}$ for $0 < ∣ y ∣ < 3$ , is equal to

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

If $cos^{- 1} (\frac{2}{3 x}) + cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4})$ , then x is equal to :

Let $S = {x \in R : 0 < x < 1 and 2 tan^{- 1} (\frac{1 - x}{1 + x}) = cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})}$ . If $n (S)$ denotes the number of elements in $S$ then :

The set of all values of $k$ for which $(tan^{- 1} x)^{3} + (cot^{- 1} x)^{3} = k π^{3}, x \in R$ , is the interval :

Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :

Let $f : [0, 4 π] \to [0, π]$ be defined by $f (x) = cos^{- 1} (cos x)$ . The number of points $x \in [0, 4 π]$ satisfying the equation $f (x) = \frac{10 - x}{10}$ is

Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then :