Question

cos2x+2sinxcosx=-1的解集怎么算的？

Answer 1

cos2x+2sinxcosx=-1
cos2x+sin2x=-1
√2cos(2x+π/4)=-1
cos(2x+π/4)=-√2/2
2x+π/4=2kπ+3π/4或2x+π/4=2kπ+5π/4
2x=2kπ+π/2或2x=2kπ+π
x=kπ+π/4或x=kπ+π/2

Answer 2

cos2x+2sinxcosx=-1
cos2x +sin2x =-1
√2 sin( 2x + π/4）= -1
sin( 2x + π/4）= -√2/2
2x +π/4 = -π/4 +2kπ
2x = -π/2 +2kπ
x = -π/4 + kπ

Answer 3

cos2x+2sinxcosx=-1
cos2x+sin2x=-1
√2sin（2x+π/4）=-1
sin（2x+π/4）=-√2/2
2x+π/4=3π/2+2kπ
解得x=5π/8+kπ k为整数

Answer 4

cos2x+2sinxcosx=-1
√2 sin( 2x + π/4）= -1 sin( 2x + π/4）= -√2/2
2x +π/4 = -π/4 +2kπ x = -π/4 + kπ k∈z