y=sinx/2+cosx/2在(-2π,2π)内的递增区间为多少？答案是-3/2π<=x<=π/2

asinx+bcosx=√(a^2+b^2)*sin(x+z)
其中tanz=b/a

所以y=√2sin(x/2+π/4)

-2π<x<2π
所以-π<x/2<π
-3π/4<x/2+π/4<5π/4

sin增区间是[2kπ-π/2,2kπ+π/2]
-3π/4<x/2+π/4<5π/4 在此范围内的是-π/2<=x/2+π/4<=π/2

-π/2<=x/2+π/4<=π/2
-3π/4<=x/2<=π/4
-3π/2<=x<=π/2

y=根号2*sin(x/2+π/4)
所以,递增的区间应该是满足:
2kπ-π/2<=x/2+π/4<=2kπ+π/2的x.
在[-2π,2π]上,就是:
[-2π,-3π/2],[-π/2,π/2],[3π/2,2π].