The number of the solutions of the equation `3sinx + 4cosx-x^2-16 = 0` is

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the equation \(3 \sin x + 4 \cos x - x^2 - 16 = 0\) and determine the number of solutions, we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate the trigonometric part:
\[ 3 \sin x + 4 \cos x = x^2 + 16 \]

### Step 2: Finding the Range of the Left Side
The expression \(3 \sin x + 4 \cos x\) can be analyzed using the properties of sine and cosine. The maximum and minimum values of \(a \sin x + b \cos x\) can be found using the formula:
\[ \text{Range} = [-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}] \]
For our case, \(a = 3\) and \(b = 4\):
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Thus, the range of \(3 \sin x + 4 \cos x\) is:
\[ [-5, 5] \]

### Step 3: Analyzing the Right Side
Next, we analyze the right side of the equation, \(x^2 + 16\). Since \(x^2\) is always non-negative, the minimum value of \(x^2 + 16\) occurs when \(x = 0\):
\[ x^2 + 16 \geq 16 \]
Thus, the right side is always greater than or equal to 16.

### Step 4: Comparing Both Sides
Now we compare the ranges:
- The left side \(3 \sin x + 4 \cos x\) has a maximum value of 5.
- The right side \(x^2 + 16\) has a minimum value of 16.

Since the maximum value of the left side (5) is less than the minimum value of the right side (16), there are no points where these two sides can be equal.

### Conclusion
Therefore, there are no solutions to the equation \(3 \sin x + 4 \cos x - x^2 - 16 = 0\).

The number of solutions is:
\[ \boxed{0} \]