The equation `sinx+sin2x+2sinxsin2x=2cosx+cos2x` is satisfied by values of `x` for which

Options

A. `x = npi +(-1)^(n) pi/6, n int I`
B. `x=2npi + (2pi)/3, n int I`
C. `x = 2npi- (2pi)/3, n int I`
D. `x = 2npi -pi/2, n int I`

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin x + \sin 2x + 2 \sin x \sin 2x = 2 \cos x + \cos 2x \]
We can express \( \sin 2x \) and \( \cos 2x \) in terms of \( \sin x \) and \( \cos x \):
\[ \sin 2x = 2 \sin x \cos x \quad \text{and} \quad \cos 2x = 1 - 2 \sin^2 x \]
Substituting these into the equation gives:
\[ \sin x + 2(2 \sin x \cos x) + 2 \sin x (2 \sin x \cos x) = 2 \cos x + (1 - 2 \sin^2 x) \]

### Step 2: Simplify the equation
Now simplify the left-hand side:
\[ \sin x + 2 \sin x \cos x + 4 \sin^2 x \cos x = 2 \cos x + 1 - 2 \sin^2 x \]
Rearranging gives:
\[ \sin x + 2 \sin x \cos x + 4 \sin^2 x \cos x + 2 \sin^2 x - 2 \cos x - 1 = 0 \]

### Step 3: Factor out common terms
Factor out \( \sin x \):
\[ \sin x (1 + 2 \cos x + 4 \sin x \cos x) + (2 \sin^2 x - 2 \cos x - 1) = 0 \]

### Step 4: Set factors to zero
This gives us two equations to solve:
1. \( \sin x = 0 \)
2. \( 1 + 2 \cos x + 4 \sin x \cos x + 2 \sin^2 x - 2 \cos x - 1 = 0 \)

### Step 5: Solve \( \sin x = 0 \)
The solutions for \( \sin x = 0 \) are:
\[ x = n\pi, \quad n \in \mathbb{Z} \]

### Step 6: Solve the second equation
Now we simplify the second equation:
\[ 2 \sin^2 x + 2 \cos x + 4 \sin x \cos x - 2 \cos x = 0 \]
This simplifies to:
\[ 2 \sin^2 x + 4 \sin x \cos x = 0 \]
Factoring gives:
\[ 2 \sin x (\sin x + 2 \cos x) = 0 \]
Setting each factor to zero gives:
1. \( \sin x = 0 \) (already solved)
2. \( \sin x + 2 \cos x = 0 \) or \( \sin x = -2 \cos x \)

### Step 7: Solve \( \sin x = -2 \cos x \)
Dividing both sides by \( \cos x \) (where \( \cos x \neq 0 \)):
\[ \tan x = -2 \]
The general solution for this is:
\[ x = \tan^{-1}(-2) + n\pi, \quad n \in \mathbb{Z} \]

### Step 8: Combine solutions
Combining all solutions, we have:
1. \( x = n\pi, \quad n \in \mathbb{Z} \)
2. \( x = \tan^{-1}(-2) + n\pi, \quad n \in \mathbb{Z} \)

### Conclusion
The equation \( \sin x + \sin 2x + 2 \sin x \sin 2x = 2 \cos x + \cos 2x \) is satisfied by values of \( x \) for which:
\[ x = n\pi \quad \text{or} \quad x = \tan^{-1}(-2) + n\pi, \quad n \in \mathbb{Z} \]