The number of solutions of the equation `tan x sin x-1=tanx-sinx, AA in [0, 2pi]` is euqal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

### Step 1: Rewrite the equation
Start by rewriting the equation in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x} \]
Substituting this into the equation gives:
\[ \frac{\sin x}{\cos x} \sin x - 1 = \frac{\sin x}{\cos x} - \sin x \]
This simplifies to:
\[ \frac{\sin^2 x}{\cos x} - 1 = \frac{\sin x - \sin x \cos x}{\cos x} \]

### Step 2: Clear the denominator
Multiply through by \( \cos x \) (noting that \( \cos x \neq 0 \)):
\[ \sin^2 x - \cos x = \sin x - \sin x \cos x \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ \sin^2 x - \sin x + \sin x \cos x - \cos x = 0 \]
This can be factored or rearranged further if needed.

### Step 4: Factor out common terms
We can factor out \( \sin x \):
\[ \sin x (\sin x - 1 + \cos x) = \cos x \]
This leads us to:
\[ \sin x - 1 = \frac{\cos x}{\sin x - 1} \]

### Step 5: Set up the equation for \( \tan x \)
From the rearrangement, we can derive:
\[ -\sin x = \cos x \]
This implies:
\[ \tan x = -1 \]

### Step 6: Solve for \( x \)
The solutions for \( \tan x = -1 \) occur at:
\[ x = \frac{3\pi}{4} + n\pi \]
For \( n = 0 \) and \( n = 1 \):
- \( x = \frac{3\pi}{4} \) (in the interval \( [0, 2\pi] \))
- \( x = \frac{7\pi}{4} \) (in the interval \( [0, 2\pi] \))

### Step 7: Count the solutions
Thus, there are two solutions in the interval \( [0, 2\pi] \):
1. \( x = \frac{3\pi}{4} \)
2. \( x = \frac{7\pi}{4} \)

### Final Answer
The number of solutions of the equation in the interval \( [0, 2\pi] \) is **2**.

---