Solve: 2tan²x + sec²x = 2, for 0 ≤ x ≤ 2π

Solve: 2tan²x + sec²x = 2, for 0 ≤ x ≤ 2π

2tan²x + sec²x = 2

which gives tan x = ± 1 /√3

If we take tan = 1 /√3, then x = π/6 or 7π/6

Again, if we take tanx = −1 /√3, then x = 5π/6 or 11π/6

Therefore, the possible solutions of above equations are

x = π/6, 5π/6, 7π/6, 11π/6 (for 0 ≤ x ≤ 2π)