38. cos4x+2cos²x=
cos2*2x+2cos²x=0
cos²2x-sin²2x+2cos²x=0
(cos²x-sin²x)²-(2sinxcosx)²+2cos²x=0
(cos²x-1+cos²x)²-4sin²xcos²x+2cos²x=0
(2cos²x-1)²-4(1-cos²x)cos²x+2cos²x=0
4cos⁴x-4cos²x+1-4(cos²x-cos⁴x)+2cos²x=0
4cos⁴x-4cos²x+1-4cos²x+4cos⁴x+2cos²x=0
8cos⁴x-6cos²x+1=0
Пусть cos²x=y
8y²-6y+1=0
D=36-32=4
y₁=6-2= 1
16 4
y₂=6+2= 1
16 2
При у= 1
4
cos²x= 1
4
cos²x - 1 =0
4
(cosx-1)(cosx+1)=0
2 2
cosx- 1 =0 cosx+1 =0
2 2
cosx= 1 cosx=-1
2 2
x=+ arccos 1 + 2πn x=+ arccos(-1) +2πn
2 2
x=+ π + 2πn x=+ (π - π )+2πn
3 3
x=+2π +2πn
3
При у= 1
2
cos²x= 1
2
cos²x - 1 =0
2
(cosx-1) (cosx+1)=0
√2 √2
cosx - 1 =0 cosx+ 1 =0
√2 √2
cosx= 1 cosx= - 1
√2 √2
x=+ arccos 1 + 2πn x=+ arccos(-1) +2πn
√2 √2
x=+ π + 2πn x=+ (π - π) + 2πn
4 4
x=+3π + 2πn
4
Ответ: х=+ π + 2πn
4
x=+3π + 2πn
4
x=+ π + 2πn
3
x=+2π +2πn
3