Solve the general vlaue.
`tan theta + 4 cot 2 theta + 1=0`

Answer

To solve the equation \( \tan \theta + 4 \cot 2\theta + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \tan \theta + 4 \cot 2\theta + 1 = 0 \]
We can express \( \cot 2\theta \) in terms of \( \tan \theta \). Recall that:
\[ \cot 2\theta = \frac{1}{\tan 2\theta} \]
Using the double angle identity for tangent:
\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
Thus,
\[ \cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta} \]
Substituting this into our equation gives:
\[ \tan \theta + 4 \cdot \frac{1 - \tan^2 \theta}{2 \tan \theta} + 1 = 0 \]

### Step 2: Simplify the Equation
Now, simplify the equation:
\[ \tan \theta + \frac{4(1 - \tan^2 \theta)}{2 \tan \theta} + 1 = 0 \]
This simplifies to:
\[ \tan \theta + \frac{2(1 - \tan^2 \theta)}{\tan \theta} + 1 = 0 \]
Multiplying through by \( \tan \theta \) to eliminate the fraction (assuming \( \tan \theta \neq 0 \)):
\[ \tan^2 \theta + 2(1 - \tan^2 \theta) + \tan \theta = 0 \]
This expands to:
\[ \tan^2 \theta + 2 - 2\tan^2 \theta + \tan \theta = 0 \]
Combining like terms gives:
\[ -\tan^2 \theta + \tan \theta + 2 = 0 \]
Rearranging, we have:
\[ \tan^2 \theta - \tan \theta - 2 = 0 \]

### Step 3: Solve the Quadratic Equation
Now we can solve the quadratic equation \( \tan^2 \theta - \tan \theta - 2 = 0 \) using the quadratic formula:
\[ \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -1, c = -2 \):
\[ \tan \theta = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \tan \theta = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \]
This gives us two solutions:
\[ \tan \theta = \frac{4}{2} = 2 \quad \text{and} \quad \tan \theta = \frac{-2}{2} = -1 \]

### Step 4: Find General Solutions
Now we find the general solutions for \( \theta \):
1. For \( \tan \theta = 2 \):
\[ \theta = \tan^{-1}(2) + n\pi, \quad n \in \mathbb{Z} \]
2. For \( \tan \theta = -1 \):
\[ \theta = -\frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the general solutions are:
\[ \theta = n\pi + \tan^{-1}(2), \quad n \in \mathbb{Z} \]
\[ \theta = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]