Решите уравнение:
1) Sin 4x Cos2x = Sin 2x Cos 4x
2) Cos 2x - Sin x = 0
3) Cos (0,5п-2x) + Sin x = 0

1)sin4xcos2s-sin2xcos4x=0sin(4x-2x)=0⇒sin2x=0⇒2x=πn⇒x=πn/22)1-2sin²x-sinx=02sin²x+sinx-1=0sinx=a⇒2a²+a-1=0D=1+8=9a1=(-1-3)/4=-1⇒sinx=-1⇒x=-π/2+2πna2=(-1+3)/4=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn3)sin2x+sinx=02sinxcosx+sinx=0sinx(2cosx+1)=0sinx=0⇒x=πncosx=-1/2⇒x=+-2π/3+2πn