If `sin A+cos2A=1//2` and `cos A+sin2A=1//3`. Then find the value of `sin 3A`.

Explanation

To solve the problem, we will follow these steps:

Given:
1. \( \sin A + \cos 2A = \frac{1}{2} \) (Equation 1)
2. \( \cos A + \sin 2A = \frac{1}{3} \) (Equation 2)

### Step 1: Square both equations
We will square both equations and then add them together.

\[ (\sin A + \cos 2A)^2 + (\cos A + \sin 2A)^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 \]

### Step 2: Expand the squares
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \):

\[ \sin^2 A + \cos^2 2A + 2\sin A \cos 2A + \cos^2 A + \sin^2 2A + 2\cos A \sin 2A = \frac{1}{4} + \frac{1}{9} \]

### Step 3: Simplify the right-hand side
Finding a common denominator for \( \frac{1}{4} \) and \( \frac{1}{9} \):

\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
\[ \Rightarrow \frac{1}{4} + \frac{1}{9} = \frac{9 + 4}{36} = \frac{13}{36} \]

### Step 4: Use Pythagorean identities
We know that \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 2A + \cos^2 2A = 1 \):

\[ 1 + 1 + 2(\sin A \cos 2A + \cos A \sin 2A) = \frac{13}{36} \]

### Step 5: Combine terms
This simplifies to:

\[ 2 + 2(\sin A \cos 2A + \cos A \sin 2A) = \frac{13}{36} \]

### Step 6: Isolate the sine term
Subtract 2 from both sides:

\[ 2(\sin A \cos 2A + \cos A \sin 2A) = \frac{13}{36} - 2 \]

Convert 2 to a fraction with a denominator of 36:

\[ 2 = \frac{72}{36} \Rightarrow \frac{13}{36} - \frac{72}{36} = \frac{13 - 72}{36} = \frac{-59}{36} \]

### Step 7: Divide by 2
Now divide both sides by 2:

\[ \sin A \cos 2A + \cos A \sin 2A = \frac{-59}{72} \]

### Step 8: Use the sine addition formula
Recognizing that \( \sin A \cos 2A + \cos A \sin 2A = \sin(A + 2A) = \sin(3A) \):

\[ \sin(3A) = \frac{-59}{72} \]

### Final Answer
Thus, the value of \( \sin 3A \) is:

\[ \sin 3A = \frac{-59}{72} \]