Question

Solve the equation on the interval 0≤x < 2.
2
4 tan²x - 12 = 0

Answer 1

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 / √