Trig Question, Help Needed Fast

Question

find x where 0 ≤ x ≤ 2pi.

cos x - tan^2 x=1.

give your answers as a fraction ordered from least to greatest.

Answer 1

cos(x)-tan²(x)=1

1: add tan²(x) on both sides, cos(x)-tan²(x)+tan²(x) = 1+tan²(x)

2: trigonometric identity, cos(x) = 1+tan²(x) = sec²(x)

3: divide by sec²(x) on both sides, cos(x)/sec²(x) = sec²(x)/sec²(x) = 1

4: sec(x)=1/cos(x), cos(x)/sec²(x) = cos(x)/(1/cos²(x)) = cos(x)(cos²(x)) = cos³(x)=1

5: cube root both sides, (cos³(x))^1/3 = (1)^1/3 => cos(x) = 1

6: solve for x, cos(x) = 1 when x=0+/-2pi(n), where n is an integer

Answer 2

cos x - tan² x = 1

cos x = 1 + tan² x

cos x = sec² x

cos x = (1 / cos x)²

cos x = 1 / cos² x

cos³ x = 1

cox x = 1

For what values of x is this true?