To solve the equation
\[
\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1,
\]
we will follow these steps:
### Step 1: Simplify the first term
The expression \(\sin^3 \theta - \cos^3 \theta\) can be factored using the identity for the difference of cubes:
\[
\sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta).
\]
Thus, we can simplify the first term:
\[
\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} = \sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta.
\]
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we can further simplify:
\[
\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta = 1 + \sin \theta \cos \theta.
\]
### Step 2: Simplify the second term
Next, we simplify the second term:
\[
\sqrt{1 + \cot^2 \theta} = \sqrt{\csc^2 \theta} = |\csc \theta|.
\]
Thus, the second term becomes:
\[
-\frac{\cos \theta}{|\csc \theta|} = -\cos \theta \cdot \sin \theta = -\sin \theta \cos \theta.
\]
### Step 3: Simplify the third term
The third term is:
\[
-2 \tan \theta \cot \theta = -2.
\]
### Step 4: Combine all terms
Now we can combine all the simplified terms:
\[
1 + \sin \theta \cos \theta - \sin \theta \cos \theta - 2 = -1.
\]
This simplifies to:
\[
1 - 2 = -1.
\]
### Step 5: Solve the equation
The equation holds true, which means we need to find the values of \(\theta\) in the interval \([0, 2\pi]\) where the conditions are satisfied.
### Step 6: Analyze the conditions
1. **Condition for \(\sin \theta\) and \(\cos \theta\)**:
- \(\sin \theta > 0\) implies \(\theta\) is in the first or second quadrant.
- \(\sin \theta - \cos \theta \neq 0\) implies \(\theta \neq \frac{\pi}{4}\) and \(\theta \neq \frac{5\pi}{4}\).
- \(\cos \theta \neq 0\) implies \(\theta \neq \frac{\pi}{2}\) and \(\theta \neq \frac{3\pi}{2}\).
### Step 7: Determine the valid intervals
From the above conditions, the valid intervals for \(\theta\) are:
- First quadrant: \((0, \frac{\pi}{4})\)
- Second quadrant: \((\frac{\pi}{4}, \frac{\pi}{2})\)
Thus, the solution for \(\theta\) is in the intervals:
\[
\theta \in (0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi).
\]
### Final Answer
The solution set for \(\theta\) in the interval \([0, 2\pi]\) is:
\[
\theta \in (0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi).
\]
