The number of distinct real roots of the equation `sqrt(sin x)-(1)/(sqrt(sin x))=cos x("where" 0le x le 2pi)` is

Options

A. 1
B. 2
C. 3
D. more than 3

Explanation

To find the number of distinct real roots of the equation

\[ \sqrt{\sin x} - \frac{1}{\sqrt{\sin x}} = \cos x \]

for \(0 \leq x \leq 2\pi\), we will follow these steps:

### Step 1: Rewrite the equation
First, we rewrite the given equation:

\[ \sqrt{\sin x} - \frac{1}{\sqrt{\sin x}} = \cos x \]

### Step 2: Square both sides
Next, we square both sides to eliminate the square root:

\[ \left(\sqrt{\sin x} - \frac{1}{\sqrt{\sin x}}\right)^2 = \cos^2 x \]

Expanding the left side:

\[ \sin x - 2 + \frac{1}{\sin x} = \cos^2 x \]

### Step 3: Use the Pythagorean identity
Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can rewrite the equation:

\[ \sin x - 2 + \frac{1}{\sin x} = 1 - \sin^2 x \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ \sin^2 x + \sin x - 3 + \frac{1}{\sin x} = 0 \]

### Step 5: Multiply through by \(\sin x\)
To eliminate the fraction, we multiply through by \(\sin x\) (noting that \(\sin x \neq 0\)):

\[ \sin^3 x + \sin^2 x - 3\sin x + 1 = 0 \]

### Step 6: Factor the polynomial
We can test for rational roots. By substituting \(\sin x = 1\):

\[ 1 + 1 - 3 + 1 = 0 \]

So, \(\sin x = 1\) is a root. We can factor the polynomial as:

\[ (\sin x - 1)(\sin^2 x + 2\sin x - 1) = 0 \]

### Step 7: Solve the quadratic equation
Now we solve the quadratic equation:

\[ \sin^2 x + 2\sin x - 1 = 0 \]

Using the quadratic formula:

\[ \sin x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]

### Step 8: Determine valid solutions
Thus, we have:

1. \(\sin x = 1\)
2. \(\sin x = -1 + \sqrt{2}\)

Next, we check if \(-1 + \sqrt{2}\) is within the valid range for \(\sin x\):

\[ -1 + \sqrt{2} \approx 0.414 \]

This value is valid since it lies between -1 and 1.

### Step 9: Find the angles
1. For \(\sin x = 1\):
- \(x = \frac{\pi}{2}\)

2. For \(\sin x = -1 + \sqrt{2}\):
- This gives two solutions in the range \(0 \leq x \leq 2\pi\):
- \(x = \arcsin(-1 + \sqrt{2})\)
- \(x = \pi - \arcsin(-1 + \sqrt{2})\)

### Conclusion
Thus, we have three distinct real roots:

1. \(x = \frac{\pi}{2}\)
2. \(x = \arcsin(-1 + \sqrt{2})\)
3. \(x = \pi - \arcsin(-1 + \sqrt{2})\)

Therefore, the number of distinct real roots of the equation is **3**.

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