Понизим степень
(1+cos(6x-π/2))²/4+(1+cos6x)²/4=1/4
(1+sin6x)²+(1+cos6x)²=1
1+2sin6x+sin²6x+1+2cos6x+cos²6x=1
2sin6x+2cos6x=1-3
sin6x+cos6x=-1
2sin3xcos3x+cos²3x-sin²3x+sin²3x+cos²3x=0
2cos²3x+2sin3xcos3x=0
2cos3x(cos3x+sin3x)=0
cos3x=0⇒3x=π/2+πk⇒x=π/6+πk/3,k∈z
sin3x+cos3x=0/cos3x
tg3x+1=0⇒tg3x=-1⇒3x=-π/4+πk⇒x=-π/12+πk/3,k∈z