58.5 Sin[(X-20)/2]=10 SinX Cos20,求X是多少?

3个回答

  • 郭敦顒回答:

    用尝试—逐步逼近法在解实际较复杂应用方程方面很有效,本人在实际工作中常用此法,也用此法解答了不少网友所提难题.现仍用此法求解所给方程——

    ∵58.5 Sin[(X-20)/2]=10 SinX Cos20

    ∴58.5 Sin[(X-20)/2]=10 SinX ×0.939683

    ∴SinX=6.22544 Sin[(X-20)/2],

    设X=29°,则SinX=0.48481,

    6.22544 Sin[(X-20)/2]= 6.22544×Sin4.5°=6.22544×0.0784591=0.48844

    误差=0.48844-0.48481=0.00363,

    相对误差=0.00363/048481×100%=0.75%.

    又设X=28.9°,则SinX=0.48328,

    6.22544 Sin[(X-20)/2]= 6.22544×Sin4.45°=6.22544×0.077589=0.48303

    误差=0.48303-0.48328=-0.00025,

    相对误差=-0.00025/048328×100%=-0.052%.

    又设X=28.91°,则SinX=0.48344,

    6.22544 Sin[(X-20)/2]= 6.22544×Sin4.455°=6.22544×0.077676=0.48356

    误差=0.48356-0.48344=0.00011,

    相对误差=0.00011/048344×100%=0.023%.

    又设X=28.907°,则SinX=0.48339,

    6.22544 Sin[(X-20)/2]= 6.22544×Sin4.4535°=6.22544×0.07765=0.48341

    误差=0.48341-0.48339=0.00002,

    相对误差=0.00002/048344×100%=0.0004%.

    又设X=28.9068°,则SinX=0.48339,

    6.22544 Sin[(X-20)/2]= 6.22544×Sin4.455°=6.22544×0.077648=0.48339

    误差=0.48339-0.48339=0

    ∴X=28.9068°.