Question

4cos3x+4=0;
√3-tg2x=0;

Answer 1

1)4cos3x+4=0
4cos3x=-4
cos3x=-1
3x=π+2πk;k∈Z
x=π/3+(2/3)πk;k∈Z
2)√3-tg2x=0
tg2x=√3
2x=π/3+πk
x=π/6+(π/2)k;k∈Z