SOLUTION: Hello, I have a Trig question that wants me to find 0°<x<360° {<= less than or equal to} for sqrt2cosx = sin2x. I started by squaring both sides, but I am not sure if I wrote the

Algebra ->  Trigonometry-basics -> SOLUTION: Hello, I have a Trig question that wants me to find 0°<x<360° {<= less than or equal to} for sqrt2cosx = sin2x. I started by squaring both sides, but I am not sure if I wrote the       Log On


   

Question 966064: Hello,
I have a Trig question that wants me to find 0° K. Legters
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282%29cos%28x%29+=+sin%282x%29
sqrt%282%29cos%28x%29+=+2sin%28x%29cos%28x%29
sqrt%282%29cos%28x%29+-+2sin%28x%29cos%28x%29+=+0
cos%28x%29%2A%28sqrt%282%29+-+2sin%28x%29%29+=+0
cos(x) = 0
x = 90, 270 degs
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sqrt%282%29+-+2sin%28x%29+=+0
sin%28x%29+=+sqrt%282%29%2F2
x = 45, 135 degs