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

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) (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) (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
------------------------------------------------------
A:
Use the identity OR
Substitute the identity on the right side of the equation .

= 0 OR = 0
= 0 OR = -1
x = OR
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