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 can follow these steps:

### Step 1: Define the function
Let \( f(x) = x^3 + 3x^2 + 6x + 3 - 2\cos x \).

### Step 2: Differentiate the function
We need to find the derivative of \( f(x) \):
\[ f'(x) = 3x^2 + 6x + 6 + 2\sin x \]
Here, \( 2\sin x \) is the derivative of \( -2\cos x \).

### Step 3: Analyze the derivative
We need to analyze \( f'(x) \) in the interval \( (0, 1) \):
- The polynomial part \( 3x^2 + 6x + 6 \) is always positive for \( x \geq 0 \) because it is a quadratic function that opens upwards and has no real roots.
- The term \( 2\sin x \) is also non-negative in the interval \( (0, 1) \) since \( \sin x \) is positive in this interval.

Thus, \( f'(x) > 0 \) for all \( x \in (0, 1) \). This means that \( f(x) \) is strictly increasing in the interval \( (0, 1) \).

### Step 4: Evaluate the function at the endpoints
Next, we evaluate \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = 0^3 + 3(0^2) + 6(0) + 3 - 2\cos(0) = 3 - 2(1) = 1 \]
- At \( x = 1 \):
\[ f(1) = 1^3 + 3(1^2) + 6(1) + 3 - 2\cos(1) = 1 + 3 + 6 + 3 - 2\cos(1) = 13 - 2\cos(1) \]
Since \( \cos(1) \) is approximately \( 0.5403 \), we can estimate:
\[ f(1) \approx 13 - 2(0.5403) \approx 13 - 1.0806 \approx 11.9194 \]

### Step 5: Determine the number of solutions
Since \( f(0) = 1 > 0 \) and \( f(1) \approx 11.9194 > 0 \), and since \( f(x) \) is strictly increasing, it implies that \( f(x) \) does not cross the x-axis in the interval \( (0, 1) \). Therefore, there are no solutions in this interval.

Thus, \( n = 0 \).

### Step 6: Find \( n + 2 \)
Finally, we need to find the value of \( n + 2 \):
\[ n + 2 = 0 + 2 = 2 \]

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