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分享求解二次方程
W * cos(R) - H * sin(R) + x * cos(R) * 2 = a;
W * sin(R) + H * cos(R) + x * sin(R) * 2 = b;
已知W, H, a, b，求x和R
W * sin(R) + H * cos(R) + x * sin(R) * 2 = b;
已知W, H, a, b，求x和R
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跑马溜溜的山上 2018-07-29
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?? 这是数学题？
W * cos(R) - H * sin(R) + x * cos(R) * 2 = a;
W * sin(R) + H * cos(R) + x * sin(R) * 2 = b;
-->
W * cos(R) * sin(R) - H * sin(R) * sin(R) + x * cos(R) * sin(R) * 2 = a * sin(R);
W * sin(R) * cos(R) + H * cos(R) * cos(R) + x * sin(R) * cos(R) * 2 = b * cos(R);
-->
W * cos(R) * sin(R) - H * sin(R) * sin(R) - W * sin(R) * cos(R) - H * cos(R) * cos(R) = a * sin(R) - b * cos(R)
-->
- H = a * sin(R) - b * cos(R)
-->
a * sin(R) + H = b * cos(R)
let u = sin(R)
-->
(a*u+H)^2 = b^2*(1-u^2)
-->
a*a*u*u + H*H + 2*a*u*H = b*b - b*b*u*u
-->
(a*a+b*b)*u*u + 2*a*H*u + (H*H-b*b) = 0
let
A = (a*a+b*b)
B = 2*a*H
C = (H*H-b*b)
-->
u(0) = -B + sqrt(B*B-4*A*C)/2/A
u(1) = -B - sqrt(B*B-4*A*C)/2/A
R = asin(u[])
反带回去求取X，并验证虚解。
