The number of the solutions of the equation `2 sin^(-1) sqrt(x^(2) + x + 1) + cos^(-1) sqrt(x^(2) + x) = (3pi)/(2)` is

Options

A. 0
B. `-1`
C. 1
D. 2

Explanation

To solve the equation \( 2 \sin^{-1}(\sqrt{x^2 + x + 1}) + \cos^{-1}(\sqrt{x^2 + x}) = \frac{3\pi}{2} \), we will follow these steps:

### Step 1: Determine the domain of the functions involved
The expressions inside the inverse sine and cosine functions must be valid.

1. For \( \sqrt{x^2 + x + 1} \):
- Since \( x^2 + x + 1 \) is always positive (it has no real roots), it is valid for all \( x \).
- The range of \( \sqrt{x^2 + x + 1} \) is \( [1, \infty) \) since the minimum value occurs at \( x = -\frac{1}{2} \), giving \( \sqrt{\frac{3}{4}} \).

2. For \( \sqrt{x^2 + x} \):
- This expression is valid when \( x^2 + x \geq 0 \), which factors to \( x(x + 1) \geq 0 \).
- The solutions to this inequality are \( x \leq -1 \) or \( x \geq 0 \).

### Step 2: Analyze the ranges of the inverse functions
- The range of \( \sin^{-1}(y) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) for \( y \in [-1, 1] \).
- The range of \( \cos^{-1}(y) \) is \( [0, \pi] \) for \( y \in [0, 1] \).

### Step 3: Set up the equation
The equation can be rewritten as:
\[ 2 \sin^{-1}(\sqrt{x^2 + x + 1}) + \cos^{-1}(\sqrt{x^2 + x}) = \frac{3\pi}{2} \]
Let \( y_1 = \sqrt{x^2 + x + 1} \) and \( y_2 = \sqrt{x^2 + x} \).

### Step 4: Solve for specific values of \( x \)
1. **Case 1: \( x = -1 \)**
- \( y_1 = \sqrt{(-1)^2 + (-1) + 1} = \sqrt{1} = 1 \)
- \( y_2 = \sqrt{(-1)^2 + (-1)} = \sqrt{0} = 0 \)
- LHS: \( 2 \sin^{-1}(1) + \cos^{-1}(0) = 2 \cdot \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2} \) (valid solution)

2. **Case 2: \( x = 0 \)**
- \( y_1 = \sqrt{0^2 + 0 + 1} = \sqrt{1} = 1 \)
- \( y_2 = \sqrt{0^2 + 0} = \sqrt{0} = 0 \)
- LHS: \( 2 \sin^{-1}(1) + \cos^{-1}(0) = 2 \cdot \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2} \) (valid solution)

### Step 5: Conclusion
Both \( x = -1 \) and \( x = 0 \) satisfy the equation, hence the number of solutions is **2**.