Solve the equation ` a cos x + cot x +1="cosec " x `

Answer

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

### Step 1: Rewrite the equation in terms of sine and cosine
We 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: Multiply through by \(\sin x\) to eliminate the denominators
Multiplying the entire equation by \(\sin x\) (assuming \(\sin x \neq 0\)):
\[ a \cos x \sin x + \cos x + \sin x = 1 \]

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

### Step 4: Square both sides
Now, we square both sides:
\[ (\sin x + \cos x)^2 = (1 - a \cos x \sin x)^2 \]
Expanding both sides:
\[ \sin^2 x + 2 \sin x \cos x + \cos^2 x = 1 - 2a \cos x \sin x + a^2 \cos^2 x \sin^2 x \]
Using \(\sin^2 x + \cos^2 x = 1\):
\[ 1 + 2 \sin x \cos x = 1 - 2a \cos x \sin x + a^2 \cos^2 x \sin^2 x \]

### Step 5: Simplify the equation
Cancelling 1 from both sides:
\[ 2 \sin x \cos x = -2a \cos x \sin x + a^2 \cos^2 x \sin^2 x \]
Factoring out \(\sin x \cos x\):
\[ \sin x \cos x (2 + 2a) = a^2 \cos^2 x \sin^2 x \]

### Step 6: Rearranging to form a quadratic equation
Rearranging gives:
\[ a^2 \cos^2 x \sin^2 x - (2 + 2a) \sin x \cos x = 0 \]
This is a quadratic equation in terms of \(\sin x \cos x\).

### Step 7: Solve for \(\sin x \cos x\)
Let \(y = \sin x \cos x\). The equation becomes:
\[ a^2 y^2 - (2 + 2a)y = 0 \]
Factoring out \(y\):
\[ y(a^2 y - (2 + 2a)) = 0 \]
Thus, \(y = 0\) or \(a^2 y - (2 + 2a) = 0\).

### Step 8: Solve for \(y\)
1. If \(y = 0\), then \(\sin x \cos x = 0\), which gives:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]
2. If \(a^2 y = 2 + 2a\), then:
\[ y = \frac{2 + 2a}{a^2} \]
Since \(y = \sin x \cos x = \frac{1}{2} \sin 2x\), we have:
\[ \frac{1}{2} \sin 2x = \frac{2 + 2a}{a^2} \]

### Step 9: Determine the range of \(a\)
To find the range of \(a\), we analyze the maximum and minimum values of \(\sin 2x\), which is between -1 and 1. Thus:
\[ -\frac{a^2}{2} \leq 2 + 2a \leq \frac{a^2}{2} \]
This leads to inequalities that can be solved to find the valid range of \(a\).

### Final Answer
The solutions for \(x\) are:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]
And the range of \(a\) is:
\[ a \in (-\infty, 2 - \sqrt{2}] \cup [2 + \sqrt{2}, \infty) \]