Solve `sin x tan x -sin x+ tan x-1=0` for `x in [0, 2pi]`.

Explanation

To solve the equation \( \sin x \tan x - \sin x + \tan x - 1 = 0 \) for \( x \) in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin x \tan x - \sin x + \tan x - 1 = 0 \]

### Step 2: Factor the equation
We can group the terms:
\[ \sin x \tan x - \sin x + \tan x - 1 = 0 \]
Rearranging gives:
\[ \sin x (\tan x - 1) + (\tan x - 1) = 0 \]
Now, we can factor out \( \tan x - 1 \):
\[ (\tan x - 1)(\sin x + 1) = 0 \]

### Step 3: Set each factor to zero
Now we have two factors:
1. \( \tan x - 1 = 0 \)
2. \( \sin x + 1 = 0 \)

### Step 4: Solve \( \tan x - 1 = 0 \)
Setting \( \tan x - 1 = 0 \) gives:
\[ \tan x = 1 \]
The solutions for \( \tan x = 1 \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4} \]

### Step 5: Solve \( \sin x + 1 = 0 \)
Setting \( \sin x + 1 = 0 \) gives:
\[ \sin x = -1 \]
The solution for \( \sin x = -1 \) in the interval \( [0, 2\pi] \) is:
\[ x = \frac{3\pi}{2} \]

### Step 6: Combine all solutions
Thus, the complete set of solutions for the original equation in the interval \( [0, 2\pi] \) is:
\[ x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4}, \quad x = \frac{3\pi}{2} \]

### Final Answer:
The solutions are:
\[ x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4}, \quad x = \frac{3\pi}{2} \]