SOLUTION: I need to find solutions in the interval (0,2pi)cos^2x=1-sinx

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Question 615276: I need to find solutions in the interval (0,2pi)cos^2x=1-sinx
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
cos%5E2%28x%29=1-sin%28x%29
First use cos%5E2%28x%29+=+1+-+sin%5E2%28x%29 to change the equation into an "all-sin" equation:
1+-+sin%5E2%28x%29=1-sin%28x%29

Next we solve for sin(x). This equation is quadratic in terms of sin(x) so we want one side to be zero. Subtracting 1 and adding sin%5E2%28x%29 to each side we get:
0+=+sin%5E2%28x%29+-+sin%28x%29
Factor. The GCF is sin(x):
0 = sin(x)(sin(x) - 1)
Use the Zero Product Property:
sin(x) = 0 or sin(x) - 1 = 0
Solving the second equation we get:
sin(x) = 0 or sin(x) = 1

Now that we have solved for sin(x), we can solve for x. Sin values of 0 and 1 are special angle values. So we do not need a calculator. From the special angles we know that...
For sin(x) = 0:
x+=+0+%2B+2%2Api%2An
or
x+=+pi+%2B+2%2Api%2An
For sin(x) = 1:
x+=+pi%2F2+%2B+2%2Api%2An

From these three equations we can find all the x's that are in the specified interval.
From x+=+0+%2B+2%2Api%2An...
We get no solutions. (Did you post the interval correctly? I ask because often there is a bracket, not a parenthesis, on one or maybe both ends of the interval. A bracket at one of end of an interval means "including the number at that end". A parenthesis means "not including...." Since you posted parentheses at both ends, 0 and 2pi are excluded. If either one of these is supposed to be included, then this equation says they would be a solution in the interval.)

From x+=+pi+%2B+2%2Api%2An...
we get x+=+pi as the only solution within the specified interval.

From x+=+pi%2F2+%2B+2%2Api%2An...
we get x+=+pi%2F2 as the only solution within the specified interval.