A) -2sin²x - cosx+1=0
-2(1-cos²x) -cosx+1=0
-2+2cos²x-cosx+1=0
2cos²x-cosx-1=0
Пусть cosx=y
2y²-y-1=0
D=1+8=9
y₁=1-3= -2/4 =-1/2
4
y₂=1+3=1
4
При у=-1/2
cosx=-1/2
x=+ (π- π/3) +2πn, n∈Z
x=+ 2π/3 + 2πn
При у=1
cosx=1
x=2πn, n∈Z
б) πn=1 x=-2π/3 + 2π = -2π+6π = 4π
3 3
Ответ: 4π
3