The total number of solutions of `cos x= sqrt(1- sin 2x)` in `[0, 2pi]` is equal to

Options

A. 2
B. 3
C. 5
D. none of these

Explanation

To solve the equation \( \cos x = \sqrt{1 - \sin 2x} \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x = \sqrt{1 - \sin 2x} \]
We know that \( \sin 2x = 2 \sin x \cos x \), so we can rewrite the equation as:
\[ \cos x = \sqrt{1 - 2 \sin x \cos x} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides:
\[ \cos^2 x = 1 - 2 \sin x \cos x \]
Rearranging gives:
\[ \cos^2 x + 2 \sin x \cos x - 1 = 0 \]

### Step 3: Use the Pythagorean identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can replace \( \cos^2 x \) with \( 1 - \sin^2 x \):
\[ (1 - \sin^2 x) + 2 \sin x \cos x - 1 = 0 \]
This simplifies to:
\[ -\sin^2 x + 2 \sin x \cos x = 0 \]
Factoring out \(-\sin x\):
\[ -\sin x (\sin x - 2 \cos x) = 0 \]

### Step 4: Solve for \(\sin x\)
This gives us two cases to consider:
1. \(\sin x = 0\)
2. \(\sin x - 2 \cos x = 0\) which simplifies to \(\sin x = 2 \cos x\)

### Step 5: Solve \(\sin x = 0\)
The solutions for \(\sin x = 0\) in the interval \([0, 2\pi]\) are:
\[ x = 0, \pi, 2\pi \]

### Step 6: Solve \(\sin x = 2 \cos x\)
Dividing both sides by \(\cos x\) (where \(\cos x \neq 0\)):
\[ \tan x = 2 \]
The general solution for \(\tan x = 2\) is:
\[ x = \tan^{-1}(2) + n\pi \]
In the interval \([0, 2\pi]\), we find:
1. \(x = \tan^{-1}(2)\)
2. \(x = \tan^{-1}(2) + \pi\)

### Step 7: Verify the solutions
We need to ensure that \( \tan^{-1}(2) + \pi \) is within the interval \([0, 2\pi]\). Since \( \tan^{-1}(2) \) is approximately \(1.107\), we have:
\[ \tan^{-1}(2) + \pi \approx 1.107 + 3.142 = 4.249 \]
This is within the interval \([0, 2\pi]\).

### Step 8: Count the total solutions
The solutions we have found are:
1. \(x = 0\)
2. \(x = \pi\)
3. \(x = 2\pi\)
4. \(x = \tan^{-1}(2)\)
5. \(x = \tan^{-1}(2) + \pi\)

Thus, the total number of solutions in the interval \([0, 2\pi]\) is **5**.

### Final Answer
The total number of solutions of \( \cos x = \sqrt{1 - \sin 2x} \) in \([0, 2\pi]\) is **5**.

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