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CACULUS

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THE VALUES OF X THAT ARE SOLUTIONS TO THE EQUATION [COSX]^2 = SIN2X IN THE INTERVAL 0 TO PI??

  • CACULUS - ,

    (cosx)^2 = 2sinxcox
    (cosx)^2 - 2sinxcosx = 0
    cosx(cosx - 2sinx) = 0
    cosx = 0 or cosx = 2sinx

    if cosx=0 then x = pi/2

    if cosx = 2sinx
    sinx/cosx = 1/2
    tanx = 1/2
    x = .46365

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