Let m be the number of solutions of `sin(2x)+cos(2x)+cosx+1=0` in `0 lt x lt (pi)/(2) and n = sin [tan^(-1)(tan((7pi)/(6)))+cos^(-1)(cos((7pi)/(3)))]`, then find the value of `(m^(2)+n^(2)+m+n+4)`

Answer

To solve the problem, we need to find the values of \( m \) and \( n \) based on the given equations and then compute \( m^2 + n^2 + m + n + 4 \).

### Step 1: Find \( m \)

We start with the equation:
\[ \sin(2x) + \cos(2x) + \cos(x) + 1 = 0 \]

Using the identities:
\[ \sin(2x) = 2\sin(x)\cos(x) \quad \text{and} \quad \cos(2x) = 2\cos^2(x) - 1 \]

We can rewrite the equation:
\[ 2\sin(x)\cos(x) + (2\cos^2(x) - 1) + \cos(x) + 1 = 0 \]
This simplifies to:
\[ 2\sin(x)\cos(x) + 2\cos^2(x) + \cos(x) = 0 \]

### Step 2: Factor out \( \cos(x) \)

Factoring out \( \cos(x) \):
\[ \cos(x)(2\sin(x) + 2\cos(x) + 1) = 0 \]

This gives us two cases to solve:
1. \( \cos(x) = 0 \)
2. \( 2\sin(x) + 2\cos(x) + 1 = 0 \)

### Step 3: Solve \( \cos(x) = 0 \)

In the interval \( 0 < x < \frac{\pi}{2} \), \( \cos(x) = 0 \) gives:
\[ x = \frac{\pi}{2} \]
However, this is not included in the interval, so we discard this solution.

### Step 4: Solve \( 2\sin(x) + 2\cos(x) + 1 = 0 \)

Rearranging gives:
\[ 2\sin(x) + 2\cos(x) = -1 \quad \Rightarrow \quad \sin(x) + \cos(x) = -\frac{1}{2} \]

### Step 5: Square both sides

Squaring both sides:
\[ (\sin(x) + \cos(x))^2 = \left(-\frac{1}{2}\right)^2 \]
Expanding the left side:
\[ \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{4} \]
Using \( \sin^2(x) + \cos^2(x) = 1 \):
\[ 1 + 2\sin(x)\cos(x) = \frac{1}{4} \]
This simplifies to:
\[ 2\sin(x)\cos(x) = -\frac{3}{4} \]

### Step 6: Solve for \( \sin(x) \)

Using the identity \( \sin(2x) = 2\sin(x)\cos(x) \):
\[ \sin(2x) = -\frac{3}{4} \]

### Step 7: Find solutions for \( \sin(2x) = -\frac{3}{4} \)

The equation \( \sin(2x) = -\frac{3}{4} \) has solutions in the intervals:
\[ 2x = \arcsin\left(-\frac{3}{4}\right) + 2k\pi \quad \text{and} \quad 2x = \pi - \arcsin\left(-\frac{3}{4}\right) + 2k\pi \]

Dividing by 2 gives:
\[ x = \frac{1}{2}\arcsin\left(-\frac{3}{4}\right) + k\pi \quad \text{and} \quad x = \frac{\pi}{2} - \frac{1}{2}\arcsin\left(-\frac{3}{4}\right) + k\pi \]

In the interval \( 0 < x < \frac{\pi}{2} \), we can find the number of valid solutions.

### Step 8: Count solutions

After analyzing the intervals, we find that there are **2 valid solutions** for \( m \):
\[ m = 2 \]

### Step 9: Find \( n \)

Next, we compute:
\[ n = \sin\left(\tan^{-1}\left(\tan\left(\frac{7\pi}{6}\right)\right) + \cos^{-1}\left(\cos\left(\frac{7\pi}{3}\right)\right)\right) \]

Calculating \( \tan^{-1}\left(\tan\left(\frac{7\pi}{6}\right)\right) \):
\[ \tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus:
\[ \tan^{-1}\left(\tan\left(\frac{7\pi}{6}\right)\right) = \frac{7\pi}{6} - \pi = \frac{\pi}{6} \]

Calculating \( \cos^{-1}\left(\cos\left(\frac{7\pi}{3}\right)\right) \):
\[ \frac{7\pi}{3} = 2\pi + \frac{\pi}{3} \quad \Rightarrow \quad \cos^{-1}\left(\cos\left(\frac{7\pi}{3}\right)\right) = \frac{\pi}{3} \]

So:
\[ n = \sin\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 10: Calculate final expression

Now we compute:
\[ m^2 + n^2 + m + n + 4 = 2^2 + 1^2 + 2 + 1 + 4 = 4 + 1 + 2 + 1 + 4 = 12 \]

Thus, the final answer is:
\[ \boxed{12} \]