6sin^2a=6-6cos^2a
6-6cos^2a+5cosa-7=0
-6cos^2a+5cosa-1=0
6cos^2a-5cosa+1=0
квадратное уравнение
корни
6/12=1/2 4/12=1/3
cosa=1/2
а принадлежит {2*pi*k - pi/3 ; 2*pi*k + pi/3}, k принадлежит z
cosa=1/3
a = +- arccos1/3 + 2*pi*k, k принадлежит z
1) (2(х+4))/(х-3) = (5(х-3))/(х+4)
(2х+8)/((х-3)(х-4)) - (5х-15)/((х-3)(х+4)) = 0
(2х+8-5х+15)/((х-3)(х+4)) = 0
(-3х+23)/((х-3)(х+4)) = 0
ОДЗ: (х-3)(х+4)≠0; х≠3; х≠ -4
-3х+23 = 0
-3х = -23
х = -23 : (-3)
х = 7 2/3
Cos5π/7+cos3π/4 ??
cos(π-2π/7)+cos(π-π/4)=-cos(2π/7)-cos(π/4)=
-(cos2π/7+cosπ/4)<0
(cos(2π/7)>0;cos(π/4)>0)
(cos5π/7+cos3π/4)<0