Solving trigonometry

Question

Solve for -90<x>90

tan²2x - 2sec2x +1 = 0

Answer 1

Solve for -90<x>90 sec=1/cos

tan²2x - 2sec2x +1 = 0

tan^2(2x)-2/cos(2x)=1

No answer because your working with degrees.

Answer 2

tan²(2x) - 2sec(2x) + 1 = 0

Since 1 + tan²(2x) = sec²(2x). we have -sec(2x) = 0.

So, sec(2x) = 0

2x = 90° + k(180°), where k = 0, ±1, ±2, ...

x = 45° + k(90°)

If k = 0, then x = 45°

If k = -1, then x = -45°

Other values of k give values of x that lie outside of the interval -90° < x < 90°

However, tan(2x) and sec(2x) are both undefined when x = 45° and -45°.

So, the equation has NO SOLUTION.