solve: `sinxtanx-1=tanx-sinx`

Explanation

To solve the equation \( \sin x \tan x - 1 = \tan x - \sin x \), we will follow these steps:

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

### Step 2: Substitute \(\tan x\)
Recall that \(\tan x = \frac{\sin x}{\cos x}\). We can substitute this into the equation:
\[ \sin x \left(\frac{\sin x}{\cos x}\right) - 1 = \frac{\sin x}{\cos x} - \sin x \]

### Step 3: Simplify both sides
Now we simplify both sides:
\[ \frac{\sin^2 x}{\cos x} - 1 = \frac{\sin x - \sin x \cos x}{\cos x} \]
The right side can be simplified to:
\[ \frac{\sin x (1 - \cos x)}{\cos x} \]
So the equation becomes:
\[ \frac{\sin^2 x}{\cos x} - 1 = \frac{\sin x (1 - \cos x)}{\cos x} \]

### Step 4: Eliminate the denominator
To eliminate the denominator, multiply through by \(\cos x\) (assuming \(\cos x \neq 0\)):
\[ \sin^2 x - \cos x = \sin x (1 - \cos x) \]

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

### Step 6: Use the Pythagorean identity
Using the identity \(\sin^2 x + \cos^2 x = 1\), we can replace \(\sin^2 x\):
\[ (1 - \cos^2 x) - \sin x + \cos x = 0 \]
This simplifies to:
\[ 1 - \cos^2 x - \sin x + \cos x = 0 \]

### Step 7: Rearrange again
Rearranging gives:
\[ -\cos^2 x + \cos x - \sin x + 1 = 0 \]
This is a quadratic in terms of \(\cos x\).

### Step 8: Solve for \(\cos x\)
Let \(y = \cos x\). The equation becomes:
\[ -y^2 + y - \sin x + 1 = 0 \]
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4(-1)(1 - \sin x)}}{2(-1)} \]
This simplifies to:
\[ y = \frac{1 \pm \sqrt{4\sin x}}{2} \]

### Step 9: Find values of \(x\)
Since \(\tan x = -1\), we know that:
\[ x = n\pi - \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]

### Final Solution
Thus, the general solution for \(x\) is:
\[ x = n\pi - \frac{\pi}{4} \quad \text{where } n \in \mathbb{Z} \]