Question

matlab求解三角函数方程的问题，急！！！求大神解答

实验解体继续这个数据
求解（2r）^3/V=24(sina)^3/(2-3COSa+(cosa)^3)/3.14
大神们看过来看过来啊~~~~~~~~

Answer 1

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)