Sinx (sinx+1) = 0;
sinx=0⇒x=πn,n∈z
sinx=-1⇒x=-π/2+2πk,k∈z
sin^2 x - sinx = 0;
sinx(sinx-1)=0
sinx=0⇒x=πn,n∈z
sinx=1⇒x=π/2+2πk,k∈z
tg^2 x - tgx = 0;
tgx(tgx-1)=0
tgx=0⇒x=πn,n∈z
tgx=1⇒x=π/4+πk,k∈z
ctg^2 x - ctgx =0.
ctgx(ctgx-1)=0
ctgx=0⇒x=πn,n∈z
ctgx=1⇒x=π/4+πk,k∈z