8 sec^2 x + 4 tan^2 x − 12 = 0 → sec^2x = tan^2x + 1
8 [ tan^2x + 1] + 4tan^2x - 12 = 0
8tan^2x + 8 + 4tan^2x - 12 = 0
12tan^2x - 4 = 0 divide through by 4
3tan^2x - 1 = 0
3tan^2x = 1
tan^2x = 1/3 take the square root of both sides
tanx = ±1 / √3
And on the given interval, this happens at pi/6 , 5pi/6, 7pi/6, 11pi/6
i hope this helps.
