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DeanR
I will first reveal the most obvious secret of trig: Most of the problems are 30/60/90 or 45/45/90 triangles of some sort.

[tex]\sec \theta + 2 = 0[/tex]

[tex]\dfrac{1 }{ \cos \theta} = -2[/tex]

[tex]\cos \theta = - \frac 1 2[/tex]

The rule to remember is [tex]\cos x = \cos a[/tex] has solutions

[tex]x = \pm a + 2 \pi k \quad[/tex] integer [tex]k[/tex]

Continuing where we left off, a cosine of -1/2 is a 30/60/90 triangle in the second or third quadrant; we pick second, and a little thought tells us the angle is  [tex]120^\circ[/tex].

[tex]\cos \theta = - \frac 1 2[/tex]

[tex]\cos \theta = \cos \frac {2\pi} 3[/tex]

[tex]\theta = \pm \frac {2\pi} 3 + 2 \pi k \quad[/tex] integer [tex]k[/tex]

In the range we want, that's [tex]k=0[/tex] with the plus sign and [tex]k=1[/tex] with the minus, so 

[tex]\theta = \frac {2\pi} 3[/tex] or [tex]-\frac {2\pi} 3 + 2 \pi = \frac{4 \pi} 3[/tex]

[tex]\theta = \frac {2\pi} 3[/tex] or [tex]\frac{4 \pi} 3[/tex]

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