Math  Solving Trig Equations
posted by Anonymous .
What am I doing wrong?
Equation:
sin2x = 2cos2x
Answers:
90 and 270
....
My Work:
2sin(x)cos(x) = 2cos(2x)
sin(x) cos(x) = cos(2x)
sin(x) cos(x) = 2cos^2(x)  1
cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1
cos^2(x)(1  cos^2(x)) = 4cos^4(x)  4cos^2(x) + 1
5cos^4(x)  5cos^2(x) + 1 = 0
cos(x) = (+/) 0.85065080835215
cos(x) = (+/)0.52573111211905

wouldn't it be easier to solve for 2x then divide it by two?
sin2x=2cos2x
sin2x/cos2x=2
tan 2x=2
2x=63.43 deg
x= .... you do it.
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