The general solution of the equation
`"sin" 2x+ 2 "sin" x+ 2 "cos" x + 1 = 0`is

Options

A. `3n pi - (pi)/(4), n in Z`
B. `2n pi + (pi)/(4), n in Z`
C. `2 n pi + (-1)^(n) "sin"^(-1) ((1)/(sqrt(3))), n in Z`
D. `n pi - (pi)/(4), n in Z`

Explanation

To solve the equation \( \sin 2x + 2 \sin x + 2 \cos x + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation:
\[ \sin 2x + 2 \sin x + 2 \cos x + 1 = 0 \]
We know that \( \sin 2x = 2 \sin x \cos x \), so we can substitute this into the equation:
\[ 2 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0 \]

### Step 2: Factor out common terms
Now, we can factor out 2 from the first three terms:
\[ 2 (\sin x \cos x + \sin x + \cos x) + 1 = 0 \]
This simplifies to:
\[ 2 (\sin x \cos x + \sin x + \cos x) = -1 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \sin x \cos x + \sin x + \cos x = -\frac{1}{2} \]

### Step 4: Use trigonometric identities
Next, we can express \( \sin x + \cos x \) in a more manageable form. We know:
\[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \]
Thus, we can rewrite the equation as:
\[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) + \sin x \cos x = -\frac{1}{2} \]

### Step 5: Substitute and simplify
Now we can express \( \sin x \cos x \) as:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Substituting this back into the equation gives us:
\[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) + \frac{1}{2} \sin 2x = -\frac{1}{2} \]

### Step 6: Solve for \( \sin x + \cos x \)
We can now set up the equation:
\[ \sin x + \cos x = 0 \]
This implies:
\[ \sin x = -\cos x \]
Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ \tan x = -1 \]

### Step 7: General solution
The general solution for \( \tan x = -1 \) is:
\[ x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the general solution of the equation \( \sin 2x + 2 \sin x + 2 \cos x + 1 = 0 \) is:
\[ x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

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