SOLUTION: solve 1+cosx=sinx, for [0,2pi) I was told to square both sides, but got stuck at , 1+2cosx+cosx^2=sinx^2

Algebra ->  Trigonometry-basics -> SOLUTION: solve 1+cosx=sinx, for [0,2pi) I was told to square both sides, but got stuck at , 1+2cosx+cosx^2=sinx^2      Log On


   

Question 769984: solve 1+cosx=sinx, for [0,2pi)
I was told to square both sides, but got stuck at ,
1+2cosx+cosx^2=sinx^2
Found 2 solutions by John10, reviewermath:
Answer by John10(297) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,
You did the first step is 100% correct by squaring both sides. You will have:
1 + 2Cos(x) + cos^2(x) = Sin^2(x)
Then you will use the formula: sin^2 (x) + cos^2 (x) = 1
Then sin^2 (x) = 1 - cos^2(x)
1 + 2cos(x) + cos^2(x) = 1 - cos^2(x)
2cos^2(x) + 2cos(x) = 0
2cos(x) (cos(x) + 1) = 0
cos(x) =0 OR cos(x) + 1 = 0
You can solve from here :)
John(john100185@yahoo.com)

Answer by reviewermath(1029) About Me  (Show Source):
You can put this solution on YOUR website!
Q:
solve 1+cosx=sinx, for [0,2pi)
I was told to square both sides, but got stuck at ,
1+2cosx+cosx^2=sinx^2
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A:
Use the identity cos%5E2%28x%29+%2B+sin%5E2%28x%29+=+1 OR sin%5E2%28x%29+=+1+-+cos%5E2%28x%29
Substitute the identity on the right side of the equation 1%2B2%2Acos%28x%29%2Bcos%5E2%28x%29=sin%5E2%28x%29.
1%2B2%2Acos%28x%29%2Bcos%5E2%28x%29=1+-+cos%5E2%28x%29
2%2Acos%5E2%28x%29+%2B+2%2Acos%28x%29=+0
%28cos%28x%29%29%2A%282%2Acos%28x%29+%2B+2%29=+0
cos%28x%29 = 0 OR 2%2Acos%28x%29+%2B+2 = 0
cos%28x%29 = 0 OR cos%28x%29 = -1
x = highlight%28pi%2F2%29 OR highlight%28pi%29