Suppose A and B are two angles such that `A , B in (0,pi)` and satisfy `sinA+sinB=1` and `cosA+cosB=0.` Then the value of `12cos2A+4cos2B` is____

Explanation

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ \sin A + \sin B = 1 \]
\[ \cos A + \cos B = 0 \]

2. **Express \(\cos B\)**:
From the second equation, we can express \(\cos B\) in terms of \(\cos A\):
\[ \cos B = -\cos A \]

3. **Use the Cosine Identity**:
We know that \(\cos(180^\circ - A) = -\cos A\). Therefore, we can set:
\[ B = \pi - A \]
This gives us our first equation:
\[ B = \pi - A \quad \text{(Equation 1)} \]

4. **Substitute \(B\) in the Sine Equation**:
Now, substitute \(B\) in the sine equation:
\[ \sin A + \sin(\pi - A) = 1 \]
We know that \(\sin(\pi - A) = \sin A\), so:
\[ \sin A + \sin A = 1 \]
\[ 2\sin A = 1 \]
\[ \sin A = \frac{1}{2} \]

5. **Find \(A\)**:
The value of \(A\) that satisfies \(\sin A = \frac{1}{2}\) in the interval \((0, \pi)\) is:
\[ A = 30^\circ \quad \text{or} \quad A = \frac{\pi}{6} \]

6. **Find \(B\)**:
Using Equation 1, we find \(B\):
\[ B = \pi - A = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

7. **Calculate \(12\cos 2A + 4\cos 2B\)**:
Now we need to calculate \(12\cos 2A + 4\cos 2B\):
\[ 12\cos(2 \times 30^\circ) + 4\cos(2 \times 150^\circ) \]
\[ = 12\cos(60^\circ) + 4\cos(300^\circ) \]

8. **Evaluate \(\cos(60^\circ)\) and \(\cos(300^\circ)\)**:
We know:
\[ \cos(60^\circ) = \frac{1}{2} \]
\[ \cos(300^\circ) = \cos(360^\circ - 60^\circ) = \cos(60^\circ) = \frac{1}{2} \]

9. **Substitute Values**:
Substitute these values back into the expression:
\[ = 12 \times \frac{1}{2} + 4 \times \frac{1}{2} \]
\[ = 6 + 2 = 8 \]

10. **Final Answer**:
Therefore, the value of \(12\cos 2A + 4\cos 2B\) is:
\[ \boxed{8} \]