The solution of the equation
`cos^2 theta + sintheta + 1 = 0`, lies in the interval

Options

A. `(- pi/4, pi/4)`
B. `(pi/4, (3pi)/4)`
C. `((3pi)/4, (5pi)/4)`
D. `((5pi)/4, (7pi)/4)`

Answer

To solve the equation \( \cos^2 \theta + \sin \theta + 1 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation using the Pythagorean identity
We know that \( \cos^2 \theta = 1 - \sin^2 \theta \). We can substitute this into the equation:

\[ 1 - \sin^2 \theta + \sin \theta + 1 = 0 \]

### Step 2: Simplify the equation
Combine like terms:

\[ - \sin^2 \theta + \sin \theta + 2 = 0 \]

Multiplying through by -1 gives:

\[ \sin^2 \theta - \sin \theta - 2 = 0 \]

### Step 3: Factor the quadratic equation
Now we can factor the quadratic equation:

\[ (\sin \theta - 2)(\sin \theta + 1) = 0 \]

### Step 4: Solve for \( \sin \theta \)
Setting each factor to zero gives us:

1. \( \sin \theta - 2 = 0 \) → \( \sin \theta = 2 \) (not possible, as \( \sin \theta \) must be in the range \([-1, 1]\))
2. \( \sin \theta + 1 = 0 \) → \( \sin \theta = -1 \)

### Step 5: Find the values of \( \theta \)
The value \( \sin \theta = -1 \) corresponds to:

\[ \theta = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Identify the interval
The general solution gives us \( \theta = \frac{3\pi}{2} + 2n\pi \). We need to find the specific intervals where this solution lies.

For \( n = 0 \):
\[ \theta = \frac{3\pi}{2} \approx 4.71 \text{ (in radians)} \]

For \( n = -1 \):
\[ \theta = \frac{3\pi}{2} - 2\pi = \frac{3\pi}{2} - \frac{4\pi}{2} = -\frac{\pi}{2} \text{ (not in the desired range)} \]

For \( n = 1 \):
\[ \theta = \frac{3\pi}{2} + 2\pi = \frac{3\pi}{2} + \frac{4\pi}{2} = \frac{7\pi}{2} \text{ (greater than } 2\pi\text{)} \]

Thus, the only relevant solution in the interval \( [0, 2\pi) \) is \( \theta = \frac{3\pi}{2} \).

### Conclusion
The solution of the equation \( \cos^2 \theta + \sin \theta + 1 = 0 \) lies in the interval \( \left( \frac{3\pi}{2} \right) \).