方程：cos2x+2sinxcosx=-1的解集是{x|x＝kπ+π2，x＝kπ−π4(k∈Z)}{x|x＝kπ+π2，x＝kπ−π4(k∈Z)}．

Question

方程：cos2x+2sinxcosx=-1的解集是

{x|x＝kπ+(π)/(2)，x＝kπ−(π)/(4)(k∈Z)}

{x|x＝kπ+(π)/(2)，x＝kπ−(π)/(4)(k∈Z)}

．

Answer 1

∵-1=cos2x+2sinxcosx=cos2x+sin2x═√(2)sin(2x+(π)/(4))
∴sin(2x+(π)/(4))＝−(√(2))/(2)
∴2x+(π)/(4)＝2kπ+(5π)/(4)或2x+(π)/(4)＝2kπ−(3π)/(4)
∴x=kπ+(π)/(2)或x＝kπ−(π)/(4)，k∈Z
故答案为：x=kπ+(π)/(2)或x＝kπ−(π)/(4)，k∈Z