Помогите решить 1) 3tg2 х+tgx=0 2) 4 сos2x-1=0

1)

3 tg (x) ^2 + tg (x) = 0

tg (x) [ 3 tg (x) + 1 ] = 0

tg (x) = 0 или 3 tg (x) + 1 = 0

tg (x) = 0

x = пк

3 tg (x) + 1 = 0

tg (x) = - 1/3

x = - arct (1/3) + пm

2)

4 cos (x) ^2 - 1 = 0

cos (x) ^2 = 1/4

cos (x) = (+/-) (1/2)

x = п/2 (+/-) п/6 + пк