1.1
2sin²x-5sinx+2=0
sinx=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒sinx=1/2⇒x=π/6+2πk U x=5π/6+2πk,k∈z
a2=(5+3)/4=2⇒sinx=2>1 нет решения
2.1
5sinx+3sin2x=0
5sinx+6sinxcosx=0
sinx*(5+6cosx)=0
sinx=0⇒x=πk,k∈z
5+6cosx=0⇒cosx=-5/6⇒x=π+-arccos5/6+2πk,k∈z
3.1.
cos²x-sin²x=√3/2
cos2x=√3/2
2x=+-π/6+2πk
x=+-π/12+πk,k∈z
4.1
sinx-√3cosx=0
2(1/2*sinx-√3/2*cosx)=0
2sin(x-π/3)=0
sin(x-π/3)=0
x-π/3=πk
x=π/3+πk,k∈z
5.1
cosx≤√2/2
π/4+2πk≤x≤7π/4+2πk,k∈z