$3\sin{x}-\cos{x}=0$
$\sin{2x}+2\sin{x}=2-2\cos{x}$
$3\sin{2x}+\cos{2x}-4\cos^2{x}=1$
$\tg{x}+\dfrac{1}{\cos^2{x}}=3$
$\sin{\dfrac{5x}{3}}+\cos{4x}=2$
$2\sin{2x}-\cos{2x}=\dfrac{\tg{x}+3}{\tg{x}+1}$
$\cos{2x}=\cos^2{\dfrac{3x}{2}}$
$\cos{x}\cos{3x}=-\dfrac{1}{2}$
$3\sin^3{x}+4\sin^2{x}\cos{x}-\sin{x}\cos^2{x}=2\sin{x}+3\cos{x}$
$3\cos{x}+4\sin{x}=6$
$\sin{2x}+\cos{2x}+\sin{6x}=0$