Solve `3 tan x + cot x = 5 cosec x`

Answer

To solve the equation \(3 \tan x + \cot x = 5 \csc x\), we can follow these steps:

### Step 1: Rewrite the trigonometric functions in terms of sine and cosine.
We know that:
- \(\tan x = \frac{\sin x}{\cos x}\)
- \(\cot x = \frac{\cos x}{\sin x}\)
- \(\csc x = \frac{1}{\sin x}\)

Substituting these into the equation gives:
\[ 3 \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 5 \frac{1}{\sin x} \]

### Step 2: Multiply through by \(\sin x \cos x\) to eliminate the fractions.
Multiplying both sides by \(\sin x \cos x\) results in:
\[ 3 \sin^2 x + \cos^2 x = 5 \cos x \]

### Step 3: Rearrange the equation.
Rearranging gives:
\[ 3 \sin^2 x + \cos^2 x - 5 \cos x = 0 \]

### Step 4: Use the Pythagorean identity.
Using the identity \(\sin^2 x + \cos^2 x = 1\), we can express \(\sin^2 x\) as \(1 - \cos^2 x\):
\[ 3(1 - \cos^2 x) + \cos^2 x - 5 \cos x = 0 \]
This simplifies to:
\[ 3 - 3 \cos^2 x + \cos^2 x - 5 \cos x = 0 \]
Combining like terms gives:
\[ -2 \cos^2 x - 5 \cos x + 3 = 0 \]

### Step 5: Multiply through by -1 to simplify.
Multiplying the entire equation by -1 results in:
\[ 2 \cos^2 x + 5 \cos x - 3 = 0 \]

### Step 6: Factor the quadratic equation.
To factor \(2 \cos^2 x + 5 \cos x - 3 = 0\), we can look for two numbers that multiply to \(2 \times -3 = -6\) and add to \(5\). The numbers \(6\) and \(-1\) work:
\[ 2 \cos^2 x + 6 \cos x - \cos x - 3 = 0 \]
Grouping gives:
\[ (2 \cos x - 1)(\cos x + 3) = 0 \]

### Step 7: Solve for \(\cos x\).
Setting each factor to zero gives:
1. \(2 \cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}\)
2. \(\cos x + 3 = 0 \Rightarrow \cos x = -3\) (not possible since \(\cos x\) must be in the range \([-1, 1]\))

### Step 8: Find the general solution for \(\cos x = \frac{1}{2}\).
The solutions for \(\cos x = \frac{1}{2}\) are:
\[ x = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2n\pi \]
where \(n\) is any integer.

### Final Answer:
The general solutions are:
\[ x = 2n\pi + \frac{\pi}{3} \quad \text{and} \quad x = 2n\pi - \frac{\pi}{3} \]