The equation `x^(3)+3x^(2)+6x+3-2cosx=0` has n solution(s) in (0, 1), then the value of `(n+2)` is equal to

Answer

To solve the equation \( x^3 + 3x^2 + 6x + 3 - 2\cos x = 0 \) for the number of solutions \( n \) in the interval \( (0, 1) \), we will analyze the function defined by:

\[ f(x) = x^3 + 3x^2 + 6x + 3 - 2\cos x \]

### Step 1: Differentiate the function
We first find the derivative \( f'(x) \) to determine the behavior of the function in the interval \( (0, 1) \).

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x^2) + \frac{d}{dx}(6x) + \frac{d}{dx}(3) - \frac{d}{dx}(2\cos x) \]

Calculating each term:

- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( 3x^2 \) is \( 6x \).
- The derivative of \( 6x \) is \( 6 \).
- The derivative of \( 3 \) is \( 0 \).
- The derivative of \( -2\cos x \) is \( 2\sin x \).

Thus, we have:

\[ f'(x) = 3x^2 + 6x + 6 + 2\sin x \]

### Step 2: Evaluate the derivative at the endpoints of the interval
Next, we evaluate \( f'(x) \) at \( x = 0 \) and \( x = 1 \).

1. **At \( x = 0 \)**:
\[ f'(0) = 3(0)^2 + 6(0) + 6 + 2\sin(0) = 6 > 0 \]

2. **At \( x = 1 \)**:
\[ f'(1) = 3(1)^2 + 6(1) + 6 + 2\sin(1) = 3 + 6 + 6 + 2\sin(1) > 0 \]
(Since \( \sin(1) \) is positive, \( f'(1) > 0 \).)

### Step 3: Determine the nature of the function
Since \( f'(x) > 0 \) for all \( x \in (0, 1) \), the function \( f(x) \) is strictly increasing in this interval.

### Step 4: Evaluate the function at the endpoints
Now we evaluate \( f(x) \) at the endpoints of the interval \( (0, 1) \).

1. **At \( x = 0 \)**:
\[ f(0) = 0^3 + 3(0)^2 + 6(0) + 3 - 2\cos(0) = 3 - 2(1) = 1 \]

2. **At \( x = 1 \)**:
\[ f(1) = 1^3 + 3(1)^2 + 6(1) + 3 - 2\cos(1) = 1 + 3 + 6 + 3 - 2\cos(1) \]
Since \( \cos(1) \) is positive, \( f(1) \) will be greater than \( 1 \).

### Step 5: Analyze the function's values
Since \( f(0) = 1 \) and \( f(1) > 1 \), and \( f(x) \) is strictly increasing, it follows that \( f(x) \) does not cross the x-axis in the interval \( (0, 1) \). Therefore, there are no solutions to the equation \( f(x) = 0 \) in this interval.

### Conclusion
Thus, the number of solutions \( n = 0 \). We need to find \( n + 2 \):

\[ n + 2 = 0 + 2 = 2 \]

### Final Answer
The value of \( n + 2 \) is \( \boxed{2} \).