The number of values of `x in [0, 2 pi]` that satisfy `"cot" x -"cosec"x = 2 "sin" x`, is

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the equation \( \cot x - \csc x = 2 \sin x \) for the number of values of \( x \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that:
\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \]
Substituting these into the equation gives:
\[ \frac{\cos x}{\sin x} - \frac{1}{\sin x} = 2 \sin x \]

### Step 2: Combine the left-hand side
Taking a common denominator on the left-hand side:
\[ \frac{\cos x - 1}{\sin x} = 2 \sin x \]

### Step 3: Cross-multiply
Cross-multiplying gives:
\[ \cos x - 1 = 2 \sin^2 x \]

### Step 4: Use the Pythagorean identity
We know that \( \sin^2 x = 1 - \cos^2 x \). Therefore, we can substitute:
\[ \cos x - 1 = 2(1 - \cos^2 x) \]
This simplifies to:
\[ \cos x - 1 = 2 - 2\cos^2 x \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ 2\cos^2 x + \cos x - 3 = 0 \]

### Step 6: Factor the quadratic equation
This is a quadratic equation in terms of \( \cos x \). We can factor it:
\[ (2\cos x + 3)(\cos x - 1) = 0 \]

### Step 7: Solve for \( \cos x \)
Setting each factor to zero gives us:
1. \( 2\cos x + 3 = 0 \)
\[ \cos x = -\frac{3}{2} \quad \text{(not possible since } \cos x \text{ must be in } [-1, 1]) \]
2. \( \cos x - 1 = 0 \)
\[ \cos x = 1 \]

### Step 8: Find the corresponding \( x \)
The solution \( \cos x = 1 \) corresponds to:
\[ x = 0 \quad \text{(within the interval } [0, 2\pi]) \]

### Step 9: Check for additional solutions
Since \( \cos x = 1 \) only gives us \( x = 0 \) in the interval \( [0, 2\pi] \), we check if there are any other solutions. The other factor \( 2\cos x + 3 = 0 \) does not provide any valid solutions.

### Conclusion
Thus, the only solution in the interval \( [0, 2\pi] \) is \( x = 0 \).

The number of values of \( x \) that satisfy the equation is:
\[ \boxed{1} \]