Number of solution (s) of the equations `cos^(-1) ( cos x) = x^(2)` is

Answer

To find the number of solutions for the equation \( \cos^{-1}(\cos x) = x^2 \), we can follow these steps:

### Step 1: Understand the function \( \cos^{-1}(\cos x) \)

The function \( \cos^{-1}(\cos x) \) gives us the principal value of the angle whose cosine is \( \cos x \). This means that:
\[ \cos^{-1}(\cos x) = x \quad \text{for } x \in [0, \pi] \]
and
\[ \cos^{-1}(\cos x) = 2\pi - x \quad \text{for } x \in [\pi, 2\pi] \]
However, more generally, we can express it as:
\[ \cos^{-1}(\cos x) = |x| \quad \text{for all } x \]

### Step 2: Set up the equation

Given the equation:
\[ \cos^{-1}(\cos x) = x^2 \]
we can replace \( \cos^{-1}(\cos x) \) with \( |x| \):
\[ |x| = x^2 \]

### Step 3: Analyze the cases for \( |x| \)

We need to consider two cases based on the definition of absolute value:

1. **Case 1:** \( x \geq 0 \)
- Here, \( |x| = x \), so the equation becomes:
\[ x = x^2 \]
Rearranging gives:
\[ x^2 - x = 0 \implies x(x - 1) = 0 \]
The solutions are:
\[ x = 0 \quad \text{or} \quad x = 1 \]

2. **Case 2:** \( x < 0 \)
- Here, \( |x| = -x \), so the equation becomes:
\[ -x = x^2 \]
Rearranging gives:
\[ x^2 + x = 0 \implies x(x + 1) = 0 \]
The solutions are:
\[ x = 0 \quad \text{or} \quad x = -1 \]

### Step 4: Collect all solutions

From both cases, we have the solutions:
- From Case 1: \( x = 0, 1 \)
- From Case 2: \( x = 0, -1 \)

Thus, the distinct solutions are:
\[ x = 0, 1, -1 \]

### Step 5: Count the number of distinct solutions

The distinct solutions are \( 0, 1, -1 \), which gives us a total of **3 solutions**.

### Final Answer:
The number of solutions of the equation \( \cos^{-1}(\cos x) = x^2 \) is **3**.

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