Question

помогите пожалуйста
2sin (4pi/3-x) -sin (4pi/3+x)= 0

Answer 1

2sin (4pi/3-x) -sin (4pi/3+x)= 0
-2sin(π/3-x)+sin(π/3+x)=0
-2sinπ/3cosx+2cosπ/3sinx+sinπ/3cosx+cosπ/3sinx=
=-sinπ/3cosx+3cosπ/3sinx=-√3/2*cosx+3*1/2*sinx=-√3/2cosx+3/2*sinx
------------------------------------------------
-√3/2cosx+3/2*sinx=0/cosx
3/2*tgx-√3/2=0
3/2*tgx=√3/2
tgx=-√3/3
x=-π/6+πk,k∈z

Answer 2

а откуда минус