用matlab解方程:结果里面的Z是什么?程序如下>> x=0:pi/2;>> x=solve('(32*cos(x)*sin(x))/49 - (2*sin(x))/((2 - 2*cos(x))^(1/2) + 1) + (sin(x)*(2 - 2*cos(x))^(1/2))/((2 - 2*cos(x))^(1/2) + 1)^2 = 0','x')x =2*pi*k - acos(1 - z^2/2)2*pi*k -
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用matlab解方程:结果里面的Z是什么?程序如下>> x=0:pi/2;>> x=solve('(32*cos(x)*sin(x))/49 - (2*sin(x))/((2 - 2*cos(x))^(1/2) + 1) + (sin(x)*(2 - 2*cos(x))^(1/2))/((2 - 2*cos(x))^(1/2) + 1)^2 = 0','x')x =2*pi*k - acos(1 - z^2/2)2*pi*k -
用matlab解方程:结果里面的Z是什么?程序如下
>> x=0:pi/2;
>> x=solve('(32*cos(x)*sin(x))/49 - (2*sin(x))/((2 - 2*cos(x))^(1/2) + 1) + (sin(x)*(2 - 2*cos(x))^(1/2))/((2 - 2*cos(x))^(1/2) + 1)^2 = 0','x')
x =
2*pi*k - acos(1 - z^2/2)
2*pi*k - acos(1 - z^2/2)
0
acos(1 - z^2/2) + 2*pi*k
acos(1 - z^2/2) + 2*pi*k
z是什么?k应该代表1,2,3.
用matlab解方程:结果里面的Z是什么?程序如下>> x=0:pi/2;>> x=solve('(32*cos(x)*sin(x))/49 - (2*sin(x))/((2 - 2*cos(x))^(1/2) + 1) + (sin(x)*(2 - 2*cos(x))^(1/2))/((2 - 2*cos(x))^(1/2) + 1)^2 = 0','x')x =2*pi*k - acos(1 - z^2/2)2*pi*k -
z应该是任意复数的意思吧.你可以让z等于一个复数代入x的表达式中,看看等式是否为零嘛.