求函数的单调区间:(1)y=sin(π/4-3x),(2)f(x)=sinx(sinx-cosx)

网友回答

(1)y=sin(π/4-3x)
递增2kπ-π/22kπ-3π/42kπ/3-π/4递减2kπ+π/22kπ+π/4-2kπ/3-5π/12(2)f(x)=sinx(sinx-cosx)
=sin^2x-sinxcosx
=(1-cos2x)/2-sin2x/2
=1-(sin2x+cos2x)/2
=1-根号2sin(2x+π/4)/2
递增2kπ+π/22kπ+π/4kπ+π/8递减2kπ-π/22kπ-3π/4kπ-3π/8======以下答案可供参考======
供参考答案1:
（1）将括号内的成分看成一个整体a，Y=sina这个的单调区间很清楚，再还原就行