tham khảo
Đk: cosx # 0 ↔ x # pi/2+ kpi
[ căn 2sin(x-pi/4)]^2 - 2sin^2x+ sinx/ cosx = 0
↔ (sinx - cosx)^2 . cosx - 2sin^2x. cosx + sinx = 0
↔ (1- sin2x ).cosx - 2sin^2x. cosx + sinx = 0
↔ cosx - 2sinx.cos^2x- 2sin^2x. cosx + sinx = 0
↔(cosx+ sinx) - 2sinx.cosx(cosx+sinx)= 0
↔(cosx+sinx) ( 1- sin2x ) = 0
↔ [cosx+sinx = 0
....[ 1- sin2x = 0
↔ [căn2cos(x- pi/4) = 0
....[ sin2x = 1
↔ [x= 3pi/4 + kpi ( k thuộc Z); tm
....[ x= pi/4 + kpi ( k thuôc Z) ; tm