Под г,д и под е.......

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