SOLUTION: solve: (sin squared) thetha - sin theta = (cos squared) thetha (for 0 degrees , or equal to thetha < or equal to 2 pi)

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Question 162318: solve:
(sin squared) thetha - sin theta = (cos squared) thetha (for 0 degrees , or equal to thetha < or equal to 2 pi)
Answer by adamchapman(301) About Me  (Show Source):
You can put this solution on YOUR website!
This one looks like a trig relation to me.
Remember that
cos%5E2%28theta%29%2Bsin%5E2%28theta%29=1.
This will rearrange easily to
cos%5E2%28theta%29=1-sin%5E2%28theta%29.
Put that on the right hand side of the initial equation you gave me:
sin%5E2%28theta%29-sin%28theta%29=cos%5E2%28theta%29
with our new right hand side:
sin%5E2%28theta%29-sin%28theta%29=1-sin%5E2%28theta%29.
Now we can rearrange to get zero on the right hand side:
2%2Asin%5E2%28theta%29-sin%28theta%29-1=0%29.
We treat this like a quadratic equation of the form 2x%5E2+-x+-1 where x=sin(theta)
The equation factorises to:
%282%2Asin%28theta%29%2B1%2F2%29%28sin%28theta%29-1%29=0%29.
So
2%2Asin%28theta%29%2B1%2F2=0
or
sin%28theta%29-1%29=0
keep the sin(theta on the left hand sides of these and rearrange to get
sin%28theta%29=-1%2F4
or
sin%28theta%29=1
sin%5E-1%28-1%2F4%29=-4.6083,184.6083
and
sin%5E-1%281%29=0,180,360
convert to radians by multiplying by 180%2Fpi

Hope this helps,
Adam