Question 1161423
<pre>
{{{sec^2(a)+5 = 3tan^2(a)}}}

Use 1 + tan²&theta; = sec²&theta;

{{{1+tan^2(a)+5 = 3tan^2(a)}}}

{{{6=2tan^2(theta)}}}

{{{3=tan^2(theta)}}}

{{{"" +- sqrt(3)=tan^""(theta)}}}

The ± tells us that &theta; can be 60° or &pi;/3 in all 4 quadrants.

{{{matrix(1,5,

theta,
""="",
pi/3+pi*n,
";",
2pi/3+pi*n)}}}

[Note that adding &pi;∙n takes care of &pi;/3 in all 4 quadrants,
and we don't have to add 2&pi;∙n.]

Edwin</pre>