Solution of the equation `sin (sqrt(1+sin 2 theta))= sin theta + cos theta` is `(n in Z)`

Options

A. `n pi-pi/4`
B. `n pi+pi/12`
C. `n pi+pi/6`
D. none of these

Explanation

To solve the equation \( \sin(\sqrt{1 + \sin(2\theta)}) = \sin(\theta) + \cos(\theta) \), we will follow these steps:

### Step 1: Use Trigonometric Identities
We know that:
- \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)
- \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

Using these identities, we can rewrite the left side of the equation:
\[ \sin(\sqrt{1 + \sin(2\theta)}) = \sin(\sqrt{1 + 2\sin(\theta)\cos(\theta)}) \]

### Step 2: Simplify the Expression Inside the Square Root
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can rewrite \( 1 + 2\sin(\theta)\cos(\theta) \) as:
\[ 1 + 2\sin(\theta)\cos(\theta) = \sin^2(\theta) + \cos^2(\theta) + 2\sin(\theta)\cos(\theta) = (\sin(\theta) + \cos(\theta))^2 \]
Thus, we have:
\[ \sqrt{1 + \sin(2\theta)} = \sqrt{(\sin(\theta) + \cos(\theta))^2} = |\sin(\theta) + \cos(\theta)| \]

### Step 3: Rewrite the Equation
Now, substituting back into the equation gives:
\[ \sin(|\sin(\theta) + \cos(\theta)|) = \sin(\theta) + \cos(\theta) \]

### Step 4: Set Up Cases
We need to consider two cases based on the absolute value:
1. \( \sin(\theta) + \cos(\theta) \geq 0 \)
2. \( \sin(\theta) + \cos(\theta) < 0 \)

#### Case 1: \( \sin(\theta) + \cos(\theta) \geq 0 \)
In this case, we can drop the absolute value:
\[ \sin(\sin(\theta) + \cos(\theta)) = \sin(\theta) + \cos(\theta) \]
This implies that:
\[ \sin(x) = x \quad \text{where } x = \sin(\theta) + \cos(\theta) \]
The only solution to \( \sin(x) = x \) is \( x = 0 \). Therefore:
\[ \sin(\theta) + \cos(\theta) = 0 \]
This leads to:
\[ \sin(\theta) = -\cos(\theta) \Rightarrow \tan(\theta) = -1 \]
Thus, \( \theta = -\frac{\pi}{4} + n\pi \) where \( n \in \mathbb{Z} \).

#### Case 2: \( \sin(\theta) + \cos(\theta) < 0 \)
In this case, we have:
\[ \sin(-(\sin(\theta) + \cos(\theta))) = -(\sin(\theta) + \cos(\theta)) \]
This simplifies to:
\[ \sin(x) = -x \quad \text{where } x = \sin(\theta) + \cos(\theta) \]
This case does not yield any new solutions since it leads to a contradiction.

### Final Solution
The general solution from Case 1 is:
\[ \theta = n\pi - \frac{\pi}{4} \quad \text{where } n \in \mathbb{Z} \]