Trig Q.

By given ,

3 sin² A + 2 sin² B = 1
3 sin² A = 1 - 2 sin² B
3 sin² A = cos 2Band3 sin 2A - 2 sin 2B = 0
sin 2B = (3/2) sin 2A
sin 2B = 3 sinA cosA
So we have

sin² 2B + cos² 2B = 9 sin² A cos² A + 9 sin⁴A
1 = 9 sin² A (cos² B + sin² A)
sin² A = 1/9
sin A = 1/3 or -1/3 (rejected since 0° ≤ A ≤ 90°)
Hencesin(A + 2B)
= sinA cos2B + cosA sin2B
= sinA * 3 sin² A + cosA * 3 sinA cosA
= 3 sinA (sin² A + cos² A)
= 3 sinA
= 3 (1/3)
= 1

2013-04-25 16:43:37 補充：
Sorry for the typing error :

1 = 9 sin² A (cos² B + sin² A)
should be
1 = 9 sin² A (cos² A + sin² A)