`(sin3theta)/(2cos2theta+1)=(1)/(2)` if `(n in Z)`

Options

A. `theta = 2n pi+(pi)/(6)`
B. `theta=2n pi-(pi)/(6)`
C. `theta=n pi+(-1)^(n)(pi)/(6)`
D. `theta-n pi-(pi)/(6)`

Explanation

To solve the equation \(\frac{\sin 3\theta}{2\cos 2\theta + 1} = \frac{1}{2}\), we will follow these steps:

### Step 1: Rewrite the equation
Start by cross-multiplying to eliminate the fraction:
\[ \sin 3\theta = \frac{1}{2}(2\cos 2\theta + 1) \]

### Step 2: Use trigonometric identities
Recall the identities:
- \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\)
- \(\cos 2\theta = 1 - 2\sin^2\theta\)

Substituting these identities into the equation gives:
\[ 3\sin\theta - 4\sin^3\theta = \frac{1}{2}(2(1 - 2\sin^2\theta) + 1) \]

### Step 3: Simplify the right side
Simplifying the right side:
\[ 3\sin\theta - 4\sin^3\theta = \frac{1}{2}(2 - 4\sin^2\theta + 1) = \frac{1}{2}(3 - 4\sin^2\theta) \]
This simplifies to:
\[ 3\sin\theta - 4\sin^3\theta = \frac{3}{2} - 2\sin^2\theta \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 3\sin\theta - 4\sin^3\theta + 2\sin^2\theta - \frac{3}{2} = 0 \]

### Step 5: Multiply through by 2 to eliminate the fraction
Multiply the entire equation by 2:
\[ 6\sin\theta - 8\sin^3\theta + 4\sin^2\theta - 3 = 0 \]

### Step 6: Rearrange into standard polynomial form
Rearranging gives:
\[ -8\sin^3\theta + 4\sin^2\theta + 6\sin\theta - 3 = 0 \]
or
\[ 8\sin^3\theta - 4\sin^2\theta - 6\sin\theta + 3 = 0 \]

### Step 7: Solve for \(\sin\theta\)
Let \(x = \sin\theta\). The equation becomes:
\[ 8x^3 - 4x^2 - 6x + 3 = 0 \]

### Step 8: Use the Rational Root Theorem or synthetic division
Testing possible rational roots, we find that \(x = \frac{1}{2}\) is a root.

### Step 9: Factor the polynomial
Using synthetic division, we can factor the polynomial:
\[ 8x^3 - 4x^2 - 6x + 3 = (x - \frac{1}{2})(8x^2 + 4x - 6) \]

### Step 10: Solve the quadratic equation
Now, solve \(8x^2 + 4x - 6 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 192}}{16} = \frac{-4 \pm 14}{16} \]
This gives:
\[ x = \frac{10}{16} = \frac{5}{8} \quad \text{and} \quad x = \frac{-18}{16} = -\frac{9}{8} \quad (\text{not valid since } x \text{ must be in } [-1, 1]) \]

### Step 11: Find angles for valid solutions
Thus, we have:
\[ \sin\theta = \frac{1}{2} \quad \Rightarrow \quad \theta = n\pi + (-1)^n \frac{\pi}{6} \quad (n \in \mathbb{Z}) \]

### Final Answer
The solutions for \(\theta\) are:
\[ \theta = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]