The solution of equation `(sin3x)/(2cos2x+1)=1/2` is

Options

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

Answer

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

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

### Step 2: Use trigonometric identities
Recall the identity for \(\sin 3x\):
\[ \sin 3x = 3\sin x - 4\sin^3 x \]
And for \(\cos 2x\):
\[ \cos 2x = 1 - 2\sin^2 x \]
Substituting these identities into the equation gives:
\[ 3\sin x - 4\sin^3 x = (1 - 2\sin^2 x) + \frac{1}{2} \]
This simplifies to:
\[ 3\sin x - 4\sin^3 x = \frac{3}{2} - 2\sin^2 x \]

### Step 3: Rearranging the equation
Rearranging the equation leads to:
\[ 4\sin^3 x - 3\sin x + 2\sin^2 x - \frac{3}{2} = 0 \]
To eliminate the fraction, multiply through by 2:
\[ 8\sin^3 x - 6\sin x + 4\sin^2 x - 3 = 0 \]

### Step 4: Rearranging terms
Rearranging gives:
\[ 8\sin^3 x + 4\sin^2 x - 6\sin x - 3 = 0 \]

### Step 5: Factor or use the Rational Root Theorem
To find the roots, we can try possible rational roots. Testing \( \sin x = \frac{1}{2} \):
\[ 8\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right) - 3 = 0 \]
Calculating gives:
\[ 8 \cdot \frac{1}{8} + 4 \cdot \frac{1}{4} - 3 - 3 = 1 + 1 - 3 - 3 = -4 \quad (\text{not a root}) \]
Testing \( \sin x = \frac{\sqrt{3}}{2} \):
\[ 8\left(\frac{\sqrt{3}}{2}\right)^3 + 4\left(\frac{\sqrt{3}}{2}\right)^2 - 6\left(\frac{\sqrt{3}}{2}\right) - 3 = 0 \]
Calculating gives:
\[ 8 \cdot \frac{3\sqrt{3}}{8} + 4 \cdot \frac{3}{4} - 3\sqrt{3} - 3 = 3\sqrt{3} + 3 - 3\sqrt{3} - 3 = 0 \quad (\text{is a root}) \]

### Step 6: Solve for \(x\)
Since \(\sin x = \frac{\sqrt{3}}{2}\), we find:
\[ x = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{2\pi}{3} + 2n\pi \]

### Final Solution
Thus, the general solutions are:
\[ x = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{2\pi}{3} + 2n\pi \]