Пожалуйста, решите уравнение. [latex]cosx- \sqrt{3} sinx=2cos5x[/latex]

2(1/2сosx-√3sinx)=2cos5x 2cos(x+π/3)=2cos5x cos5x-cos(x+π/3)=0 -2sin(3x+π/6)sin(2x-π/6)=0 sin(3x+π/6)=0⇒3x+π/6=πn⇒3x=-π/6+πn⇒x=-π/18+πn/3 sin(2x-π/6)=0⇒2x-π/6=πn⇒2x=π/6+πn⇒x=π/12+πn/2