方程12x²+πx-12π=0的解

方程12x²+πx-12π=0的解
方程12x²+πx-12π=0的解

方程12x²+πx-12π=0的解
12x²+πx-12π=0
x^2+πx/12=π
（x+π/24)^2=π+π^2/576=(π^2+576π）/576
x+π/24=±√(π^2+576π)/24
x1=-π/24+√ (π^2+576π)/24 x2=-π/24-√(π^2+576π)/24 .

sin(a-b)=sin(a)cos(b)-cos(a)sin(b) sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b) cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
所以三角函数等式 =(cosxcosy-sinxsiny)-(3^1/2)(sinxcosy+sinxc...

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sin(a-b)=sin(a)cos(b)-cos(a)sin(b) sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b) cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
所以三角函数等式 =(cosxcosy-sinxsiny)-(3^1/2)(sinxcosy+sinxcosy)
=cos(x+y)-(3^1/2)sin(x+y) (1)
根据一元二次方程求根公式[-b±(b^2-4ac)^(1/2)]/2a
可知 x+y={[-b+(b^2-4ac)^(1/2)]/2a }+{[-b-(b^2-4ac)^(1/2)]/2a}
=(-b/2a)+(-b/2a)=-b/a
根据一元二次方程可知 a=12 b=π 所以x+y=-π/12
所以 （1） =cos(-π/12)-(3^1/2)sin(-π/12)
=2[1/2cos(-π/12)-(1/2)(3^1/2)sin(-π/12)]
=2[sin(π/6)cos(-π/12)-cos(π/6)sin(-π/12)] （2）
根据sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
可知 （2）=2sin[(π/6)-(-π/12)]
=2sin[(2π/12)+(π/12)]
=2sin(3π/12)
=2sin(π/4)
=根号2

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-4 3带入不对啊解错了你确定题是这样？