solve the following equations if 0 <(less than or equal to) a <(less than) 2pi
2sin^2[a] = 1 - sin [a]
please answer clearly showing all working
2sin^2(a) + sin(a) - 1 = 0
Let x = sin(a). Then
2x^2 + x - 1 = 0
This factors:
(2x - 1)(x + 1) = 0
So
2x - 1 = 0 ==> x = 1/2 ==> sin(a) = 1/2
or
x + 1 = 0 ==> x = -1 ==> sin(a) = -1
So
sin(a) = 1/2 ==> a = (pi)/6 rad, 5(pi)/6 rad
or
sin(a) = -1 ==> a = 3(pi)/2 rad
Thus
a = (pi)/6, 5(pi)/6, or 3(pi)/2.
-Dan