Помогите решить не могу сделать плз прост забыл 1)[tex]cos( \frac{5 \pi }{2} +6x)cos3x=sinxcos3x[/tex] 2)[tex]cos3x-sin(7x- \frac{ \pi }{2} )=cos5x[/tex]

1cos(2π+π/2+6x)cos3x=sinxcos3xcos(π/2+6x)cos3x=sinxcos3x-sin6xcos3x=sinxcos3xsinxcos3x+sin6xcos3x)=0cos3x(sinx+sin6x)=0cos3x*2sin3,5x*cos2,5x=0cos3x=0⇒3x=π/2+πk,k∈z⇒x=π/6+πk/3,k∈zsin3,5x=0⇒3,5x=πk,k∈z⇒x=2πk/7,k∈zcos2,5x=0⇒2,5x=π/2+πk,k∈z⇒x=π/5+2πk/5,k∈z2cos3x-sin(7x-π/2)=cos5xcos3x+sin(π/2-7x)=cos5xcos3x+cos7x=cos5x2cos5xcos2x-cos5x=0cos5x(2cos2x-1)=0cos5x=0⇒5x=π/2+πk,k∈z⇒x=π/10+πk/5,k∈z2cos2x-1=0⇒cos2x=1/2⇒2x=-π/3+2πk,k∈z U 2x=π/3+2πk,k∈zx=-π/6+πk,k∈z U x=π/6+πk,k∈z