If acos x + cot x +1=cosec x then the possible values of x can be

Answer

To solve the equation \( a \cos x + \cot x + 1 = \csc x \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ a \cos x + \cot x + 1 = \csc x \]
We can express \(\cot x\) and \(\csc x\) in terms of sine and cosine:
\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \]
Substituting these into the equation gives:
\[ a \cos x + \frac{\cos x}{\sin x} + 1 = \frac{1}{\sin x} \]

### Step 2: Clear the fractions
Multiply through by \(\sin x\) to eliminate the denominators:
\[ a \cos x \sin x + \cos x + \sin x = 1 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ a \cos x \sin x + \cos x + \sin x - 1 = 0 \]

### Step 4: Group terms
We can group the terms:
\[ \cos x (a \sin x + 1) + \sin x - 1 = 0 \]

### Step 5: Isolate terms
Now isolate \(\sin x\):
\[ \sin x (1 - a \cos x) = -\cos x \]

### Step 6: Analyze the equation
This equation can have solutions based on the values of \(\sin x\) and \(\cos x\). We can set:
\[ \sin x = 0 \quad \text{or} \quad 1 - a \cos x = 0 \]

### Step 7: Solve for \(\sin x = 0\)
If \(\sin x = 0\), then:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 8: Solve for \(1 - a \cos x = 0\)
If \(1 - a \cos x = 0\), then:
\[ \cos x = \frac{1}{a} \]
This gives:
\[ x = \cos^{-1}\left(\frac{1}{a}\right) \quad \text{and} \quad x = -\cos^{-1}\left(\frac{1}{a}\right) + 2n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 9: Conclusion
The possible values of \(x\) are:
\[ x = n\pi \quad \text{and} \quad x = \cos^{-1}\left(\frac{1}{a}\right) + 2n\pi \quad \text{or} \quad x = -\cos^{-1}\left(\frac{1}{a}\right) + 2n\pi \]