Question

满足sin(x+sinx)=cos(x-cosx)的锐角x=？

Answer 1

∵0<x<π/2,
∴ 0<sinx<1，0<cosx<1
∴0<x+sinx<1+π/2，-1<x-cosx<π/2
∴0<π/2-(x-cosx)<1+π/2
∵sin(x+sinx)=cos(x-cosx)=sin[π/2-(x-cosx)]
∴x+sinx=π/2-(x-cosx)或x+sinx+π/2-(x-cosx)=π
2x+sinx+cosx=π/2或sinx+cosx=π/2(舍)
结合图象可知 x约=0.2

追问

不好意思答案为兀/4

追答

∵0<x<π/2,
∴ 0<sinx<1，0<cosx<1
∴0<x+sinx<1+π/2，-1<x-cosx<π/2
∴0<π/2-(x-cosx)<1+π/2
∵sin(x+sinx)=cos(x-cosx)=sin[π/2-(x-cosx)]
∴x+sinx=π/2-(x-cosx)或x+sinx+π/2-(x-cosx)=π
2x+sinx-cosx=π/2或sinx+cosx=π/2(舍)
注意到函数f(x)=2x+sinx-cosx在（0，π/2）是增函数，且f(π/4)=π/2
所以，满足原方程的锐角是π/4