1) sin4x+sin^2 2x=0
2SIN2X COS2X+SIN^2 2X=0
SIN2X (2COS 2X +SIN 2X)=0
SIN 2X=0 ILI 2COS 2X+SIN 2X=0
2X=πn tg 2x+2=0
x=π/2) *n tg2x=-2; 2x=-arctg2+πk; k;n-целые
Ответ. π/2) n; -arctg2+πk; n;k-целые
2)1-4sinx cosx=0
-2*(2sinx cosx)=-1
sin 2x=1/2; 2x=(-1)^n * (arcsin(1/2)+πn; 2x=(-1)^n *(π/6)+πn;
x=(-1)^n (⇵/12+⇵/2)*n
3) cos^2 x=1/2
cosx=√(1/2) ili cosx=-√(1/2)
cosx=√2 /2 x=+-(π-π/4)+2πn
x=(плюс, минус)π/4+2πn; х=+-(3π/4)+2πт
4)tgx=ctgx
tgx-1/tgx=0
(tg^2 x-1)/tgx=0; {tg^2 x-1=0; {tg^2 x=1; {tgx=1 ili tgx=-1
{tgx≠0
x=+-(π/4)+πn