Cos(2π-x)+sin(π/2+x)=√2
из следующих формулы
cos(2π-x)=cos(-x)=cosx
sin(π/2+x)=cosx
получим
cosx+cosx=√2
2cosx=√2
cosx=√2/2 ⇒ x=π/4
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3sin²(x/2)+sin(x/2)*cos(x/2)-2sin²(x/2)-2cos²(x/2)=0
sin²(x/2)+sin(x/2)cos(x/2)-2cos²(x/2)=0 (/cos²x/2≠0
(sin(x/2)/cos(x/2))²+sin(x/2)/cos(x/2)-2=0
tg²x/2+tgx/2-2=0
tgx/2=t
t²+t-2=0
t=1 t=-2
tgx/2=1 tgx/2=-2
x/2=π/4+πn x/2=-arctg2+πn
x=π/2+2πn x=-2arctg2+2πn