Question

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

Answer 1

(sinx+sin3x+sin5x)/(cosx+cos3x+cos5x) + 2tgx=0
(2sin3xcos2x+sin3x)/(2cos3xcos2x+cos3x)+2tgx=0
sin3x(2cos2x+1)/cos3x(2cos2x+1)+2tgx=0
tg3x+2tgx=0
(3tgx-tg³x)/(1-3tg²x)+2tgx=0
(3tgx-tg³x+2tgx-6tg³x)/(1-3tg²x)=0
{5tgx-7tg³x=0
{1-3tg²x≠0
tgx(5-7tg²x)=0
tgx=0⇒x=πk,k∈z
7tg²x=5⇒tg²x=5/7⇒tgx=+-√35/7⇒x=+-arctg√35/7+πk,k∈z