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 will follow these steps:

### Step 1: Rewrite the equation
The equation \( \cos x = \sqrt{1 - \sin 2x} \) can be simplified. We know that \( \sin 2x = 2 \sin x \cos x \). Therefore, 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:
\[ \cos^2 x = 1 - 2 \sin x \cos x \]

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

### Step 4: Rearrange the equation
Rearranging gives us:
\[ \sin^2 x - 2 \sin x \cos x = 0 \]

### Step 5: Factor the equation
We can factor out \( \sin x \):
\[ \sin x (\sin x - 2 \cos x) = 0 \]

### Step 6: Solve for \( \sin x = 0 \)
Setting \( \sin x = 0 \):
\[ x = 0, \pi, 2\pi \]
These are three solutions in the interval \( [0, 2\pi] \).

### Step 7: Solve for \( \sin x - 2 \cos x = 0 \)
Setting \( \sin x - 2 \cos x = 0 \) gives:
\[ \sin x = 2 \cos x \]
This can be rewritten as:
\[ \tan x = 2 \]

### Step 8: Find solutions for \( \tan x = 2 \)
The general solutions for \( \tan x = 2 \) are:
\[ x = \tan^{-1}(2) + n\pi \]
For \( n = 0 \):
\[ x = \tan^{-1}(2) \quad (\text{in } [0, \pi]) \]
For \( n = 1 \):
\[ x = \tan^{-1}(2) + \pi \quad (\text{in } [\pi, 2\pi]) \]
Thus, there are two additional solutions from this equation.

### Step 9: Total solutions
Combining the solutions, we have:
1. From \( \sin x = 0 \): \( x = 0, \pi, 2\pi \) (3 solutions)
2. From \( \tan x = 2 \): \( x = \tan^{-1}(2) \) and \( x = \tan^{-1}(2) + \pi \) (2 solutions)

Therefore, the total number of solutions in the interval \( [0, 2\pi] \) is:
\[ 3 + 2 = 5 \]

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

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