Question

求函数y＝2√2sinxcosx－√2sinx递增区间

Answer 1

y＝2√2sinxcosx－√2sinx=√2sin2x-√2sinx,y‘=√2(2cos2x-cosx),y‘=√2(2cos2x-cosx)＞0时,

√2(2cos2x-cosx)＞0,4cos²x-cosx-2＞0,当1≥cosx＞(1+√33)/8或-1≤cosx≤(1-√33)/8,递增；当(1-√33)/8＞cosx＞(1+√33)/8时,递减.