Solve the equation ` (tan x )^( cos^(2) x)= ( cot x )^( sin x )`

Answer

To solve the equation \( (\tan x)^{\cos^2 x} = (\cot x)^{\sin x} \), we will follow these steps:

### Step 1: Rewrite Cotangent
We know that \( \cot x = \frac{1}{\tan x} \). Therefore, we can rewrite the equation as:
\[ (\tan x)^{\cos^2 x} = \left(\frac{1}{\tan x}\right)^{\sin x} \]
This simplifies to:
\[ (\tan x)^{\cos^2 x} = (\tan x)^{-\sin x} \]

### Step 2: Set the Exponents Equal
Since the bases are the same, we can set the exponents equal to each other:
\[ \cos^2 x = -\sin x \]

### Step 3: Rearrange the Equation
Rearranging gives us:
\[ \cos^2 x + \sin x = 0 \]

### Step 4: Use the Pythagorean Identity
We know that \( \cos^2 x = 1 - \sin^2 x \). Substituting this into the equation gives:
\[ 1 - \sin^2 x + \sin x = 0 \]
This can be rearranged to:
\[ -\sin^2 x + \sin x + 1 = 0 \]
Multiplying through by -1 gives:
\[ \sin^2 x - \sin x - 1 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we will use the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = -1 \):
\[ \sin x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
This simplifies to:
\[ \sin x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

### Step 6: Evaluate the Solutions
Now we have two potential solutions for \( \sin x \):
1. \( \sin x = \frac{1 + \sqrt{5}}{2} \)
2. \( \sin x = \frac{1 - \sqrt{5}}{2} \)

Calculating these:
- \( \frac{1 + \sqrt{5}}{2} \) is greater than 1, which is not possible for sine.
- \( \frac{1 - \sqrt{5}}{2} \) is negative, which is also not possible for sine.

Thus, we only consider:
\[ \sin x = \frac{\sqrt{5} - 1}{2} \]

### Step 7: Find the General Solution
To find \( x \), we take the inverse sine:
\[ x = \sin^{-1}\left(\frac{\sqrt{5} - 1}{2}\right) \]
The general solution for sine is:
\[ x = n\pi + (-1)^n \sin^{-1}\left(\frac{\sqrt{5} - 1}{2}\right), \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the final solution is:
\[ x = n\pi + (-1)^n \sin^{-1}\left(\frac{\sqrt{5} - 1}{2}\right), \quad n \in \mathbb{Z} \]

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