设x,y属于[-兀/4,兀/4],a属于R且x^3+sinx-2a=0,4y^3+sinycosy+a=0,则cos(x+2y)=?

Question

设x,y属于[-兀/4,兀/4],a属于R且x^3+sinx-2a=0,4y^3+sinycosy+a=0,则
cos(x+2y)=?

Answer 1

4y^3+sinycosy+a=0两边同时乘以2,可变换为(2y)^3+sin2y+2a=0
根据x^3+sinx-2a=0,可得x^3+sinx=-((2y)^3+sin2y)=(-2y)^3+sin(-2y)
函数f(x)=x^3+sinx在题目限定的范围下是增函数
所以x=-2y,x+2y=0
cos(x+2y)=1