A)1+cosx+2cos²x-1=0
cosx+2cos²x=0
cosx(1+2cosx)=0
cosx=0⇒x=π/2+πn,n∈z
cosx=-1/2⇒x=+-2π/3+2πk,k∈z
в)1-2sin²2x+sin2x=0
sin2x=a
2a²-a-1=0
D=1+8=9
a1=(1-3)/4=-1/2⇒sin2x=-1/2⇒2x=(-1)^(n+1)*π/6+πn⇒
x=(-1)^(n+1)*π/12+πn/2,n∈z
a2=(1+3)/4=1⇒sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z
г)cosx=0⇒x=π/2+πn,n∈z
tgx=1⇒x=π/4+πk,k∈z
б)cos4x-cos(π/2-4x)=√3/2
-2sin(4x-π/4)sinπ/4=√3/2
-2*√2/2*sin(4x-π/4)=√3/2
sin(4x-π/4)=-√6/2<-1<br>нет решения