Let `f(theta)= sin theta - cos^(2) theta-1`, where `theta in R` and `m le f(theta) le M` .
Let N denotes the number of solution of the equation `f(theta)=0` in `[0,4 pi]` then the value of `log_(sqrt(m^(2)))(N)+log_(sqrt(m^(2)))((1)/(N+1))` is equal to

Options

A. `1/2`
B. 1
C. `(-1)/(2)`
D. `-1`

Answer

To solve the problem, we start by analyzing the function \( f(\theta) = \sin \theta - \cos^2 \theta - 1 \).

### Step 1: Rewrite the function
We can rewrite \( f(\theta) \) using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ f(\theta) = \sin \theta - (1 - \sin^2 \theta) - 1 \]
This simplifies to:
\[ f(\theta) = \sin \theta + \sin^2 \theta - 2 \]

### Step 2: Set the function to zero
To find the solutions, we set \( f(\theta) = 0 \):
\[ \sin \theta + \sin^2 \theta - 2 = 0 \]
Rearranging gives:
\[ \sin^2 \theta + \sin \theta - 2 = 0 \]

### Step 3: Solve the quadratic equation
Let \( x = \sin \theta \). The equation becomes:
\[ x^2 + x - 2 = 0 \]
We can factor this quadratic:
\[ (x - 1)(x + 2) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 1 \quad \text{or} \quad x = -2 \]
Since \( \sin \theta \) must be in the range \([-1, 1]\), we discard \( x = -2 \) and keep \( x = 1 \).

### Step 4: Find the angles
Now we find \( \theta \) such that \( \sin \theta = 1 \):
\[ \theta = \frac{\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Within the interval \([0, 4\pi]\), the solutions are:
\[ \theta = \frac{\pi}{2}, \quad \frac{5\pi}{2} \]
Thus, there are \( N = 2 \) solutions.

### Step 5: Determine the range of \( f(\theta) \)
Next, we need to find the minimum \( m \) and maximum \( M \) values of \( f(\theta) \).

1. **Maximum value**: When \( \sin \theta = 1 \):
\[ f\left(\frac{\pi}{2}\right) = 1 + 1 - 2 = 0 \]

2. **Minimum value**: When \( \sin \theta = -1 \):
\[ f\left(\frac{3\pi}{2}\right) = -1 + 1 - 2 = -2 \]

Thus, we have:
\[ m = -2, \quad M = 0 \]

### Step 6: Calculate the logarithmic expression
Now we need to evaluate:
\[ \log_{\sqrt{m^2}}(N) + \log_{\sqrt{m^2}}\left(\frac{1}{N+1}\right) \]
First, calculate \( m^2 \):
\[ m^2 = (-2)^2 = 4 \quad \Rightarrow \quad \sqrt{m^2} = 2 \]
Now substituting \( N = 2 \):
\[ \log_{2}(2) + \log_{2}\left(\frac{1}{3}\right) \]
Calculating each logarithm:
\[ \log_{2}(2) = 1 \]
\[ \log_{2}\left(\frac{1}{3}\right) = -\log_{2}(3) \]
Thus, we have:
\[ 1 - \log_{2}(3) \]

### Final Answer
The final expression simplifies to:
\[ 1 - \log_{2}(3) \]