(sinx+sin3x+sin5x)/(cosx+cos3x+cos5x) + 2tgx=0

(sinx+sin3x+sin5x)/(cosx+cos3x+cos5x) + 2tgx=0(2sin3xcos2c+sin3x)/(2cos3xcos2x+cos3x)+2tgx=0sin3x(2cos2x+1)/cos3x(2cos2x+1)+2tgx=0tg3x+2tgx=0(3tgx-tg³x)/(1-3tg²x)+2tgx=0 1-3tg²x≠0⇒tgx≠+-1/√3⇒x≠+-π/6+πk,k∈z (3tgx-tg³x)+2tgx(1-3tg²x)=0 3tgx-tg³x+2tgx-6tg³x=0 5tgx-7tg³x=0 tgx(5-7tg²x)=0 tgx=0⇒x=πk,k∈z 5-7tg²x=0 7tg²x=5 tg²x=5/7 tgx=-√35/7⇒x=-arctg√35/7+πk,k∈z tgx=√35/7⇒x=arctg√35/7+πk,k∈z ответ x={πk; -arctg√35/7+πk; arctg√35/7+πk,k∈z}

Раскладываем на множители sin+sin3x+sin5x sinx+sin3x+sin5x=sinx+sin(x+2x)+sin(3x+2x)=sinx+sinx*cos2x+cosx*sin2x+sin3x*cos2x+cos3x*sin2x=sinx+sinx*cos2x+2sinx*cos^2x+sin(2x+x)*cos2x+cos(x+2x)*sin2x=sinx+sinx*cos2x+2sinx*cos^2x+(2sinx*cos^2x+cos2x*sinx)*cos2x+(cosx*cos2x-sinx*sin2x)*2sinx*cosx=sinx(1+cos2x+2cos^2x+(2cos^2x+cos2x)*cos2x+2cosx*(cosx*cos2x-sinx*sin2x))=sinx(1+cos2x+2cos^2x+cos^2(2x)+2cos^2x*cos2x+2cos^2x*cos2x-4sin^2x*cos^2x)=sinx(1+cos2x+2cos^2x+cos^2(2x)+4cos^2x*cos2x-sin^2(2x))=sinx(2cos^2(2x)+cos2x+2cos^2x+4cos^2x*cos2x)=sinx(2cos^2(2x)+cos2x+1+cos2x+4cos^2x*cos2x)=sinx(2cos^2(2x)+2cos(2x)+2(1+cos2x)*cos2x+1)=sinx(2cos^2(2x)+2cos2x+2cos2x+2cos^2(2x)+1)=sinx(4cos^2(2x)+4cos(2x)+1)=sinx*(2cos(2x)+1)^2 теперь раскладываем cosx+cos3x+cos5x cosx+cos3x+cos5x=cosx+cos(2x+x)+cos(2x+3x)=cosx+cos2x*cosx-sin2x*sinx+cos2x*cos3x-sin2x*sin3x=cosx+cos2x*cosx-2sin^2x*cosx+cos2x*cos(x+2x)-2sinx*cosx*sin(x+2x)=cosx+cos2x*cosx-2sin^2x*cosx+cos2x*(cosx*cos2x-2sin^2x*cosx)-2sinx*cosx*sin(x+2x)=cosx(1+cos2x-2sin^2x+cos^2(2x)-2sin^2x*cos2x-2sinx*(sinx*cos2x+cosx*sin2x))=cosx(2cos2x+cos^2(2x)-2sin^2x*cos2x-2sin^2x*cos2x-4sin^2x*cos^2x)=cosx(2cos2x+cos^2(2x)-4sin^2x*cos2x-4sin^2x*cos^2x)=cosx(2cos2x+cos^2(2x)-2(1-cos2x)*cos2x-sin^2(2x))=cosx(2cos2x+cos^2(2x)-sin^2(2x)-2cos2x+2cos^2(2x))=cosx(2cos^2(2x)-1+2cos2x-2cos2x+2cos^2(2x))=cosx(4cos^2(2x)-1)=cosx(2cos2x-1)(2cos2x+1) подставляем в уравнение: (sinx*(2cos(2x)+1)^2)/(cosx*(2cos2x-1)(2cos2x+1))+2tgx=0 tgx*(2cos(2x)+1)/(2cos2x-1)+2tgx=0 tgx*((2cos(2x)+1)/(2cos2x-1)+2)=0 tgx=0 x1=pi*n (2cos2x+1)/(2cos2x-1)+2=0 (2cos2x+1+4cos2x-2)/(2cos2x-1)=0 (6cos2x-1)/(2cos2x-1)=0 6cos2x-1=0 cos2x=1/6 2x=arccos(1/6)+2pi*n x2=0,5arccos(1/6)+pi*n 2x=-arccos(1/6)+2pi*n x3=-0,5arccos(1/6)+pi*n ответ: x1=pi*n; x2=0,5arccos(1/6)+pi*n; x3=-0,5arccos(1/6)+pi*n