Solve equation, giving exact solutions which lie in [0, 2pi): tan2(x)=1-sec(x)

Question

I got stuck in the middle of the problem. Here's how far I got

tan²(x)=1-sec(x)

tan2(x)+sec(x)-1=0

sin²(x)/cos²(x)+1/cos(x)-1=0

Answer 1

tan²x = 1 - secx but tan²x + 1 = sec²x So

sec²x - 1 = 1 - secx or sec²x + secx -2 = 0

Now, let secx = z then, z² + z -2 = 0 Now solve for z = secx using the quadratic equation!

You can take it from here!