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matlab求解三角函数方程的问题,
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求解(2r)^3/V=24(sina)^3/(2-3COSa+(cosa)^3)/3.14
人气:450 ℃ 时间:2020-05-26 03:14:09
解答
已知V、r,求r
syms V a,solve('(2*r)^3/V=24*(sin(a))^3/(2-3*cos(a)+(cos(a))^3)/3.14')
r1=0.98491/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)
r2= -0.49246/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)+.85296*i/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)
r2=-0.49246/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)-.85296*i/sin(a)/(3.+sin(a)^2)*(V*(2.*cos(a)+sin(a)^2*cos(a)+2.)*sin(a)^2*(3.+sin(a)^2)^2)^(1/3)
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