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 :)

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- April 11th 2007, 04:16 AMT_o_Msolving trigonometry quadradic
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 :) - April 11th 2007, 06:03 AMtopsquark
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