Because you cannot divide by,
!!!!
Because it might be zero!
Then you divided by zero!
(Oh no!)
Once you divide by zero you get strange results.
That was you fault.
-----
Instead:
Thus,
Factor,
Thus,
Now set each factor equal to zero.
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?
Thank you, The Perfect Hacker and Captain Black. Interestingly, with this particular equation, the value that did not show up when I divided by (1 + Sin(theta)) (1 - Sin(theta)) was 3PI/2, which in fact is the value that creates the divide by zero situation for this equation ... (1 + Sin(3PI/2)).
Ed