Question

$log_{sin3x}(cosx-cos2x)=1$ Задача повышенной сложности, 11 класс.

Answer 1

по свойству логарифма
cos(x)-cos(2x)=sin(3x)

ОДЗ sin(3x)>0 sin(3x)<>1  
ОДЗ  (0;π/6)U(π/6;π/3)U(2π/3;5π/6)U(5π/6;π)U(4π/3;3π/2)U(3π/2;5π/3)

2sin(3x/2)*sin(x/2)=2sin(3x/2)*cos(3x/2)

sin(3x/2)=0
3x/2=πn
x=2πn/3  Вне ОДЗ

sin(x/2)=cos(3x/2)
sin(x/2)-sin(π/2-3x/2)=0
2sin(x-π/4)*cos(π/4-x/2)=0

cos(π/4-x/2)=0
x=π/2+2πn  Вне ОДЗ

sin(x-π/4)=0
x=π/4+πn       5π/4+2πn Вне ОДЗ

Ответ : x=π/4+2πn