1) cosx<–½ 2) sin>½ 3) 6sin²x–4sin²x–1

Question

1) cosx<–½<br> 2) sin>½
3) 6sin²x–4sin²x–1

Answer 1

6sin ^ 2x-4sin2x +1 = 0 
6sin ^ 2x-8sinxcosx + (sin^2x+cos^2x) = 0 
7sin ^ 2x-8sinxcosx + соs^2x = 0  - delim na cos^2x
7tg^2x-8tgx+1=0
Замена: tgx=a
7a^2-8a+1=0
a1=1;a2=1/7
Reshaem 2 uravneniya:
1)tgx=1
x=arctg1+Pin=Pi/4+Pin; 
2)tgx=1/7
x=arctg1/7+Pin; (n принадлежит Z)