已知f(x)=sin2x+3sinx+3cosx(0

问题描述:

已知f(x)=sin2x+3sinx+3cosx(0

f'(x)=2cos2x+3cosx-3sinx
=2(cosx+sinx)(cosx-sinx)+3(cosx-sinx)
=(cosx-sinx)(2cosx+2sinx+3)
=-√2sin(x-π/4)[2√2sin(x+π/4)+3]
因为-2√20
所以就看-√2sin(x-π/4)符号
-π/4