Find the number of solution of the equation
`x^(3)+2x+cosx+tanx=0 in (-(pi)/(2),(pi)/(2))`

Explanation

To find the number of solutions for the equation \( x^3 + 2x + \cos x + \tan x = 0 \) in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we can follow these steps:

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

### Step 2: Find the derivative
To analyze the behavior of the function, we will compute its derivative:
\[ f'(x) = 3x^2 + 2 - \sin x + \sec^2 x \]
Here, \( \sec^2 x = \frac{1}{\cos^2 x} \) is the derivative of \( \tan x \).

### Step 3: Analyze the derivative
We need to determine whether \( f'(x) \) is always positive in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \):
- The term \( 3x^2 \) is always non-negative.
- The term \( 2 \) is a constant positive value.
- The term \( -\sin x \) varies from -1 to 1.
- The term \( \sec^2 x \) is always positive and tends to infinity as \( x \) approaches \( \pm \frac{\pi}{2} \).

Thus, we can conclude that:
\[ f'(x) = 3x^2 + 2 - \sin x + \sec^2 x > 0 \]
for all \( x \) in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). This implies that \( f(x) \) is a monotonically increasing function.

### Step 4: Evaluate the function at the endpoints
Next, we evaluate \( f(x) \) at the endpoints of the interval:
- As \( x \) approaches \( -\frac{\pi}{2} \):
\[ f\left(-\frac{\pi}{2}\right) = \left(-\frac{\pi}{2}\right)^3 + 2\left(-\frac{\pi}{2}\right) + \cos\left(-\frac{\pi}{2}\right) + \tan\left(-\frac{\pi}{2}\right) \to -\infty \]
because \( \tan\left(-\frac{\pi}{2}\right) \) tends to \( -\infty \).

- As \( x \) approaches \( \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 + 2\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) + \tan\left(\frac{\pi}{2}\right) \to +\infty \]
because \( \tan\left(\frac{\pi}{2}\right) \) tends to \( +\infty \).

### Step 5: Conclusion
Since \( f(x) \) is continuous and monotonically increasing, and it transitions from \( -\infty \) to \( +\infty \) over the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), by the Intermediate Value Theorem, there exists exactly one \( x \) in the interval such that \( f(x) = 0 \).

Thus, the number of solutions to the equation \( x^3 + 2x + \cos x + \tan x = 0 \) in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) is **1**.

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