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Math - Solving Trig Equations

posted by on .

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

  • Math - Solving Trig Equations - ,

    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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