2)
Пусть cosx = t
D = 9 - 4*2*(-2) = 25t1 = (3+5)/4 = 2t2 = (3-5)/4 = -1/2cosx = 2x= +- arccos2 + 2*pi*n, n€Zcosx = -1/2x = +- (pi - arccos1/2) + 2*pi*n, n€Z3) ctgx + 3tgx = 2 sqrt (3)1/tgx + tgx = 2sqrt (3)(tg^2x + 1)/tgx = 2sqrt (3)tg^x - 2sqrt (3)*tgx +1 =0Пусть tgx=tt^2-2sqrt (3)t+1 =0D = 12 - 4*1 = 8t1,2 = (2sqrt (3)+-2sqrt (2))/2x = arctg(sqrt (3)+sqrt (2)) + pi*n, n €Zx= arctg (sqrt (3)-sqrt (2)) + pi*n, n €Z4) tgx = ctgx tgx/ctgx = 1tg^2x = 1tgx=1tgx=-1x=pi/4 + pi*n, n €Zx= -pi/4 + pi*n, n €Z5 и 6 не уверен, как решать.