Find the general solution of `"cosec"x=-2 and cot x =sqrt(3)`

Answer

To find the general solution for the equations \( \csc x = -2 \) and \( \cot x = \sqrt{3} \), we will solve each equation step by step.

### Step 1: Solve \( \csc x = -2 \)

The cosecant function is the reciprocal of the sine function:
\[ \csc x = \frac{1}{\sin x} \]
Thus, we can rewrite the equation:
\[ \sin x = -\frac{1}{2} \]

### Step 2: Find the principal value of \( x \)

The sine function equals \(-\frac{1}{2}\) at specific angles in the third and fourth quadrants. The reference angle for \(\sin x = \frac{1}{2}\) is \(\frac{\pi}{6}\). Therefore, the angles where \(\sin x = -\frac{1}{2}\) are:
\[ x = \frac{7\pi}{6}, \quad x = \frac{11\pi}{6} \]

### Step 3: Write the general solution for \( \csc x = -2 \)

The general solution for \(\csc x = -2\) can be expressed as:
\[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]

### Step 4: Solve \( \cot x = \sqrt{3} \)

The cotangent function is the reciprocal of the tangent function:
\[ \cot x = \frac{1}{\tan x} \]
Thus, we can rewrite the equation:
\[ \tan x = \frac{1}{\sqrt{3}} \]

### Step 5: Find the principal value of \( x \)

The tangent function equals \(\frac{1}{\sqrt{3}}\) at specific angles in the first and third quadrants. The reference angle for \(\tan x = \frac{1}{\sqrt{3}}\) is \(\frac{\pi}{6}\). Therefore, the angles where \(\tan x = \frac{1}{\sqrt{3}}\) are:
\[ x = \frac{\pi}{6}, \quad x = \frac{7\pi}{6} \]

### Step 6: Write the general solution for \( \cot x = \sqrt{3} \)

The general solution for \(\cot x = \sqrt{3}\) can be expressed as:
\[ x = n\pi + \frac{\pi}{6}, \quad n \in \mathbb{Z} \]

### Step 7: Combine the solutions

Now we have two general solutions:
1. From \( \csc x = -2 \):
\[ x = n\pi + (-1)^n \frac{\pi}{6} \]
2. From \( \cot x = \sqrt{3} \):
\[ x = n\pi + \frac{\pi}{6} \]

### Final General Solution

The general solutions for the equations \( \csc x = -2 \) and \( \cot x = \sqrt{3} \) are:
- For \( \csc x = -2 \):
\[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]
- For \( \cot x = \sqrt{3} \):
\[ x = n\pi + \frac{\pi}{6}, \quad n \in \mathbb{Z} \]