`(dy)/(dx) ((1+ cos x)/(y)) =- sin x and f ((pi)/(2)) =-1,` then `f (0)` is:

Options

A. -2
B. 1
C. 3
D. 4

Answer

To solve the given differential equation and find the value of \( f(0) \), we will follow these steps:

### Step 1: Write the differential equation
The given differential equation is:
\[ \frac{dy}{dx} \cdot \frac{1 + \cos x}{y} = -\sin x \]

### Step 2: Separate the variables
We can separate the variables by rearranging the equation:
\[ \frac{dy}{y} = -\frac{\sin x}{1 + \cos x} dx \]

### Step 3: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{dy}{y} = \int -\frac{\sin x}{1 + \cos x} dx \]

The left side integrates to:
\[ \ln |y| + C_1 \]

### Step 4: Simplify the right side integral
For the right side, we can use the substitution \( t = 1 + \cos x \). Then, the derivative is:
\[ dt = -\sin x \, dx \quad \Rightarrow \quad -\sin x \, dx = dt \]
Thus, the integral becomes:
\[ \int -\frac{\sin x}{1 + \cos x} dx = \int \frac{dt}{t} = \ln |t| + C_2 \]
Substituting back for \( t \):
\[ \ln |1 + \cos x| + C_2 \]

### Step 5: Combine the results
Now we equate the two integrals:
\[ \ln |y| = \ln |1 + \cos x| + C \]
where \( C = C_2 - C_1 \).

### Step 6: Exponentiate both sides
Exponentiating both sides gives:
\[ |y| = e^C |1 + \cos x| \]
Let \( k = e^C \), then:
\[ y = k(1 + \cos x) \]

### Step 7: Use the initial condition
We are given that \( f\left(\frac{\pi}{2}\right) = -1 \). Substituting \( x = \frac{\pi}{2} \):
\[ -1 = k(1 + \cos(\frac{\pi}{2})) = k(1 + 0) = k \]
Thus, \( k = -1 \).

### Step 8: Write the final function
Now we can write the function:
\[ y = -1(1 + \cos x) = -1 - \cos x \]

### Step 9: Find \( f(0) \)
Now we need to find \( f(0) \):
\[ f(0) = -1 - \cos(0) = -1 - 1 = -2 \]

### Final Answer
Thus, the value of \( f(0) \) is:
\[ \boxed{-2} \]