Can I get help with this pre-calculus problem?

Question

I need to find all solutions (0≤x≤2pie) of
2sec^2 x + tan^2 x -3 =0
if you guys can show work that would be great.

Answer 1

2sec^2x + tan^2x = 3
2(1/cos^2x) + sin^2x/cos^2x = 3
(2+sin^2x) / cos^2x = 3
(3-(1-sin^2x))/cos^2x = 3
(3-cos^2x/cos^2x = 3
3/cos^2x - 1 = 3
3/cos^2x = 4
cos^2x = 3/4
cosx = +/-sqrt(3/4) = +/-sqrt(3)/2
so answers are Pi/6 5Pi/6 7Pi/6 and 11Pi/6

Answer 2

sec² = 1 + tan², so

2sec² x + tan² x - 3 = 0
2(1 + tan² x) + tan² x - 3 = 0
3tan² x - 1 = 0
tan² x = 1/3
tan x = √3 / 3
x = π/6 or 7π/6