This can be written:
1/cos2(x)+cos2(x)/sin2(x)=13/3.
Let y=cos(x), then sin(x)=1-y2:
1/y2+y2/(1-y2)=13/3,
1-y2+y4=13y2(1-y2)/3,
3-3y2+3y4=13y2-13y4,
16y4-16y2+3=0=(4y2-3)(4y2-1).
4y2=3, y=±√3/2; 4y2-1, y=±½.
Therefore, cos(x)=±½,±√3/2.
x=π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6, etc.
More generally: x=(6n±1)π/6, (3n±1)π/3, where n is an integer.