This is the equation:

1 + Sin(theta) = 2 Cos squared(theta)

The method I used goes like this:

1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities

1 + Sin(theta)

------------------------------ = 2 divide by [1 - Sin squared(theta)]

(1 + Sin(theta)) (1 - Sin(theta))

1

--------------- = 2

(1 - Sin(theta))

1 - Sin(theta) = 1/2

-Sin(theta) = -1/2

Sin(theta) = 1/2

theta = PI/6 + 2k(PI) or theta = 5PI/6 + 2k(PI)

restrict the solution to 0 < theta < 2PI

therefore the solution is {PI/6, 5PI/6}

HOWEVER you get a MORE complete solution if you work via this method:

1 + Sin(theta) = 2 Cos squared(theta) original equation

1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities

1 + Sin(theta) = 2 - 2 Sin squared(theta) multiplies out the right side

2 Sin squared(theta) + Sin(theta) - 1 = 0 rearranges the equation

(2 Sin(theta) - 1) (Sin(theta) + 1) = 0 factors left side

2 Sin(theta) - 1 = 0 or Sin(theta) + 1 = 0

Sin(theta) = 1/2 or Sin(theta) = -1

theta = {PI/6, 5PI/6} or theta = 3PI/2

Clearly the second method yields a more complete answer, and I verified it with a TI-89. Yet the first method is algebraicly sound, as far as I can tell. Any thoughts on this?