Sin3xsin^3 x+cos3xcos^3 x=1/8

$sin3x=sin(x+2x)=sinx~cos2x+cosx~sin2x=\\sinx(cos^2x-sin^2x)+2sinx~cos^2x=\\sinx~cos^2x-sin^3x+2sinx~cos^2x=3sinx~cos^2x-sin^3x\\\\ cos3x=cos(x+2x)=cosx~cos2x-sinx~sin2x=\\cosx(cos^2x-sin^2x)-2sin^2x~cosx=cos^3x-3cosx~sin^2x\\\\ sin3x~sin^3x+cos3x~cos^3x=\\(3sinx~cos^2x-sin^3x)sin^3x+(cos^3x-3cosx~sin^2x)cos^3x=\\-sin^6x+3sin^4x~cos^2x-3cos^4x~sin^2x+cos^6x=\\(cos^2x-sin^2x)^3=cos^32x\\\\cos^32x=1/8=(1/2)^3\\cos2x=1/2\\2x=arccos(1/2)+2\pik=\pi/3+2\pik\\x=\pi/6+pik$

Проверка:
x=π/6
$x=\pi/6\\sin3x~sin^3x+cos3x~cos^3x=\\sin(3*\pi/6)sin^3(\pi/6)+cos(3*\pi/6)cos^3(\pi/6)=\\sin(\pi/2)(1/2)^3+cos(\pi/2)cos^3(\pi/6)=1*(1/8)+0*cos^3(\pi/6)=1/8$