Question

4sin(2pix)*cos^2(2pix)+1+sin(4pix)+2cos(2pix)=0

Answer 1

$4sin(2\pi x)cos^2(2\pi x)+1+sin(4\pi x)+2cos(2\pi x)=0\\\\cos(2\pi x)\cdot (2sin(2\pi x)cos(2\pi x)+1)+(1+sin(4\pi x))=0\\\\cos(2\pi x)\cdot (sin(4\pi x)+1)+(1+sin(4\pi x))=0\\\\(1+sin(4\pi x))(cos(2\pi x)+1)=0\\\\1)\; sin(4\pi x)=-1,\; 4\pi x=-\frac{\pi}{2}+2\pi n,\; n\in Z\\\\x=-\frac{1}{8}+\frac{n}{2},\; n\in Z\\\\2)\; cos(2\pi x)=-1,\; 2\pi x=\pi +2\pi k,\; k\in Z\\\\x=\frac{1}{2}+k,\; k\in Z$