Cos5x+cos7x=cos(π+6x) так как cos(α+π)=-cos(α) то
cos5x+cos7x=-cos6x
cos5x+cos6x+cos7x=0
так как cosα+cosβ=2cos(α+β)/2 *cos(α-β)/2 то получаем
(cos7x+cos5x)+cos6x=0
2cos((5+7)/2)*cos((7-5)/2)+cos6x=0
2cos6x*cosx+cos6x=0
cos6x(2cosx+1)=0
cos6x=0
6x=π/2+πn n=0,1,2,3․․․․
x=π/12+πn n=0,1,2,3․․․․
или 2cosx+1=0
cosx=-1/2
x=2π/3+2πn n=0,1,2,3․․․․
x=4π/3+2πn n=0,1,2,3․․․․
ответы так же
x=π/12+πn n=0,1,2,3․․․․
x=2π/3+2πn n=0,1,2,3․․․․
x=4π/3+2πn n=0,1,2,3․․․․