已知x,y∈[-π/4 ,π/4 ],a∈R,且 x^3+sinx-2a=0,4y^3+(1/2)sin2y+a=0.求cos(x+2y)的值.原方程组化为x^3+sinx=2a 以及 (-2y)^3+sin(-2y)=2a .∵当x,-2y∈[-π/2 ,π/2]时,函数f(t)=t^3+sint在[-π/2 ,π/2]上单调递增,又 f(x)=f(-2y)∴x=
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![已知x,y∈[-π/4 ,π/4 ],a∈R,且 x^3+sinx-2a=0,4y^3+(1/2)sin2y+a=0.求cos(x+2y)的值.原方程组化为x^3+sinx=2a 以及 (-2y)^3+sin(-2y)=2a .∵当x,-2y∈[-π/2 ,π/2]时,函数f(t)=t^3+sint在[-π/2 ,π/2]上单调递增,又 f(x)=f(-2y)∴x=](/uploads/image/z/8840693-29-3.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%E2%88%88%5B-%CF%80%2F4+%2C%CF%80%2F4+%5D%2Ca%E2%88%88R%2C%E4%B8%94+x%5E3%2Bsinx-2a%3D0%2C4y%5E3%2B%281%2F2%29sin2y%2Ba%3D0.%E6%B1%82cos%28x%2B2y%29%E7%9A%84%E5%80%BC.%E5%8E%9F%E6%96%B9%E7%A8%8B%E7%BB%84%E5%8C%96%E4%B8%BAx%5E3%2Bsinx%3D2a+%E4%BB%A5%E5%8F%8A+%28-2y%29%5E3%2Bsin%28-2y%29%3D2a+.%E2%88%B5%E5%BD%93x%2C-2y%E2%88%88%5B-%CF%80%2F2+%2C%CF%80%2F2%5D%E6%97%B6%2C%E5%87%BD%E6%95%B0f%28t%29%3Dt%5E3%2Bsint%E5%9C%A8%5B-%CF%80%2F2+%2C%CF%80%2F2%5D%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E5%8F%88+f%28x%29%3Df%28-2y%29%E2%88%B4x%3D)
已知x,y∈[-π/4 ,π/4 ],a∈R,且 x^3+sinx-2a=0,4y^3+(1/2)sin2y+a=0.求cos(x+2y)的值.原方程组化为x^3+sinx=2a 以及 (-2y)^3+sin(-2y)=2a .∵当x,-2y∈[-π/2 ,π/2]时,函数f(t)=t^3+sint在[-π/2 ,π/2]上单调递增,又 f(x)=f(-2y)∴x=
已知x,y∈[-π/4 ,π/4 ],a∈R,且 x^3+sinx-2a=0,4y^3+(1/2)sin2y+a=0.求cos(x+2y)的值.
原方程组化为x^3+sinx=2a 以及 (-2y)^3+sin(-2y)=2a .
∵当x,-2y∈[-π/2 ,π/2]时,函数f(t)=t^3+sint在[-π/2 ,π/2]上单调递增,
又 f(x)=f(-2y)
∴x=-2y,∴cos(x+2y)=cos0=1.
疑问:为什么要证单调性
已知x,y∈[-π/4 ,π/4 ],a∈R,且 x^3+sinx-2a=0,4y^3+(1/2)sin2y+a=0.求cos(x+2y)的值.原方程组化为x^3+sinx=2a 以及 (-2y)^3+sin(-2y)=2a .∵当x,-2y∈[-π/2 ,π/2]时,函数f(t)=t^3+sint在[-π/2 ,π/2]上单调递增,又 f(x)=f(-2y)∴x=
因为f(t)=t^3+sint-2a是单调函数,所以只存在一个t'使得f(t')=t‘^3+sint’-2a=0,故如果 f(x)=f(-2y)
说明x=-2y
f(x)=f(-2y),如果不是单调递增或单调递减的话,是不可能得到x=-2y的。你自已画图就知道了,如果f(x)在这个区间有很多递增及单调递减区间,就有很多解,不能得到x=-2y