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March 26, 2017

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2sinxcosx+4sin^2xcos^2x=0
solve for x in radians between [0,2pi)

(I mean that it is sine squared x and cosine squared x not sine to the power of 2x or cosine to the power of 2x)

  • trig - ,

    factor out sin*cos to get

    2sin*cos(1+2sin*cos) = 0
    so,
    sin=0: x = 0 or π
    or
    cos=0: x = π/2 or 3π/2
    or
    1+sin2x = 0
    sin2x = -1
    2x = 3π/2: x = 3π/4

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