The number of values of `x in [-2pi, 2pi]` which satisfy the equation `"cosec x"=1+cot x` is equal to

Options

A. 0
B. 2
C. 4
D. 6

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \csc x = 1 + \cot x \]
Recall that \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \). Therefore, we can rewrite the equation as:
\[ \frac{1}{\sin x} = 1 + \frac{\cos x}{\sin x} \]

### Step 2: Combine the terms on the right-hand side
To combine the terms on the right-hand side, we can find a common denominator (which is \( \sin x \)):
\[ \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x} \]

### Step 3: Eliminate the denominator
Since \( \sin x \neq 0 \) (to avoid division by zero), we can multiply both sides by \( \sin x \):
\[ 1 = \sin x + \cos x \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \sin x + \cos x = 1 \]

### Step 5: Use a trigonometric identity
We can use the identity \( \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \):
\[ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = 1 \]

### Step 6: Solve for \( \sin\left(x + \frac{\pi}{4}\right) \)
Dividing both sides by \( \sqrt{2} \):
\[ \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 7: Find the general solutions
The general solutions for \( \sin A = \frac{1}{\sqrt{2}} \) are:
\[ A = \frac{\pi}{4} + 2n\pi \quad \text{and} \quad A = \frac{3\pi}{4} + 2n\pi \]
Substituting back for \( A \):
\[ x + \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi \quad \Rightarrow \quad x = 2n\pi \]
\[ x + \frac{\pi}{4} = \frac{3\pi}{4} + 2n\pi \quad \Rightarrow \quad x = \frac{3\pi}{4} - \frac{\pi}{4} + 2n\pi = \frac{\pi}{2} + 2n\pi \]

### Step 8: Find specific solutions within the interval
Now we need to find the values of \( n \) such that \( x \) lies within \( [-2\pi, 2\pi] \).

1. For \( x = 2n\pi \):
- \( n = -1 \) gives \( x = -2\pi \)
- \( n = 0 \) gives \( x = 0 \) (not valid since \( \sin x = 0 \))
- \( n = 1 \) gives \( x = 2\pi \) (not valid since \( \sin x = 0 \))

2. For \( x = \frac{\pi}{2} + 2n\pi \):
- \( n = -1 \) gives \( x = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2} \)
- \( n = 0 \) gives \( x = \frac{\pi}{2} \)
- \( n = 1 \) gives \( x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2} \) (not valid since it exceeds \( 2\pi \))

### Step 9: Valid solutions
The valid solutions we found are:
- \( x = -2\pi \)
- \( x = -\frac{3\pi}{2} \)
- \( x = \frac{\pi}{2} \)

### Conclusion
Thus, the number of values of \( x \) in the interval \( [-2\pi, 2\pi] \) that satisfy the equation \( \csc x = 1 + \cot x \) is **3**.