√(tgx(1 + cosx)) + √(tgx(1 - cosx)) - 2√tgx · cosx = 0
1 + cosx > 0 при х∈R
1 - cosx ≥ 0 при x∈R
ОДЗ: tgx ≥ 0
√tgx · (√(1 + cosx) + √(1 - cosx) - 2cosx) = 0
1) tgx = 0
x = πn
2) √(1 + cosx) + √(1 - cosx) - 2cosx = 0
1/2 cos(x/2) + 1/2 sin(x/2) - 2(cos²(x/2) - sin²(x/2)) = 0
1/2 (cos(x/2) + sin(x/2) ) - 2(cos(x/2) + sin(x/2))(cos(x/2) - sin(x/2)) = 0
(cos(x/2) + sin(x/2)) · (1/2 - 2cos(x/2) + 2sin(x/2)) = 0
a) cos(x/2) + sin(x/2) = 0
tg(x/2) = -1
x/2 = - π/4 + πk
x = - π/2 + 2πk - не подходит, т.к. в этой точке тангенс не определен
б) 1/2 - 2cos(x/2) + 2sin(x/2) = 0
cos(x/2) - sin(x/2) = 1/4
1/√2 cos(x/2) - 1/√2 sin(x/2) = 1/(4√2)
cos(x/2 + π/4) = 1/(4√2)
x/2 + π/4 = + - arccos(1/(4√2)) + 2πm
x/2 = + - arccos(1/(4√2)) - π/4 + 2πm
x = + - 2arccos(1/(4√2)) - π/2 + 4πm
х = - 2arccos(1/(4√2)) - π/2 + 4πm не подходит, т.к. tgx<0, <br>Ответ: πn, х = 2arccos(1/(4√2)) - π/2 + 4πm