Г
(4-5cosx-2+2cos²x)/√sinx=0
{sinx≠0
{2cos²x-5cosx+2=0
cosx=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πk,k∈z
a2=(5=30/4=2⇒cosx=2>1 нет решения
д
2sin3x*cos2x=cos2x
2sin3xcos2x-cos2x=0
cos2x*(2sin3x-1)=0
[cosx=0⇒x=π/2+πk,k∈z
[sin3x=1/2⇒3x=π/6+2πk U 3x=5π/6+2πk⇒
⇒x=π/18+2πk/3 U x=5π/18+2πk/3,k∈z
е
2cos(x-6π+π/2)*cosx=sinx
2cos(x+π/2)*cosx=sinx
-2sinx*cosx-sinx=0
-sinx*(2cosx+1)=0
[sinx=0⇒x=πk,k∈z
[cosx=-1/2⇒x=+-2π/3+2πk,k∈z