Number of solutions of the equation `cot^(-1)sqrt(4-x^(2))+cos^(-1)(x^(2)-5)=(3pi)/2` is:

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the equation \( \cot^{-1}(\sqrt{4 - x^2}) + \cos^{-1}(x^2 - 5) = \frac{3\pi}{2} \), we will analyze the domains and ranges of the functions involved and find the possible values of \( x \).

### Step 1: Determine the domain of \( \sqrt{4 - x^2} \)

The expression inside the square root must be non-negative:
\[ 4 - x^2 \geq 0 \]
This simplifies to:
\[ x^2 \leq 4 \]
Thus, we have:
\[ -2 \leq x \leq 2 \]

**Hint:** Always check the conditions for square roots to ensure the expression is defined.

### Step 2: Determine the domain of \( \cos^{-1}(x^2 - 5) \)

The argument of the \( \cos^{-1} \) function must lie within the interval \([-1, 1]\):
\[ -1 \leq x^2 - 5 \leq 1 \]
This gives us two inequalities to solve:
1. \( x^2 - 5 \geq -1 \) which simplifies to \( x^2 \geq 4 \) or \( |x| \geq 2 \)
2. \( x^2 - 5 \leq 1 \) which simplifies to \( x^2 \leq 6 \) or \( |x| \leq \sqrt{6} \)

Combining these two results, we find:
\[ |x| \geq 2 \quad \text{and} \quad |x| \leq \sqrt{6} \]
Thus, the valid intervals for \( x \) are:
\[ x \in [-\sqrt{6}, -2] \cup [2, \sqrt{6}] \]

**Hint:** When dealing with inverse trigonometric functions, always ensure the inputs are within their defined ranges.

### Step 3: Combine the domains

From Step 1, we have \( -2 \leq x \leq 2 \) and from Step 2, we have \( x \in [-\sqrt{6}, -2] \cup [2, \sqrt{6}] \).

The intersection of these intervals gives us:
\[ x = -2 \quad \text{and} \quad x = 2 \]

**Hint:** Always look for intersections of intervals when combining conditions.

### Step 4: Check the solutions in the original equation

1. **For \( x = 2 \)**:
\[ \cot^{-1}(\sqrt{4 - 2^2}) + \cos^{-1}(2^2 - 5) = \cot^{-1}(0) + \cos^{-1}(-1) \]
\[ = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \]
This satisfies the equation.

2. **For \( x = -2 \)**:
\[ \cot^{-1}(\sqrt{4 - (-2)^2}) + \cos^{-1}((-2)^2 - 5) = \cot^{-1}(0) + \cos^{-1}(-1) \]
\[ = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \]
This also satisfies the equation.

### Conclusion

Both \( x = 2 \) and \( x = -2 \) are valid solutions. Therefore, the number of solutions to the equation is:

\[ \boxed{2} \]