## anonymous 5 years ago Solve: sec²θ = 2tanθ+4 over the interval [0,2pi)

1. Yuki

sec^2 = 1 + tan^2 so if you substitute that you will get $\tan^2(x) -2\tan(x) -3 = 0$

2. Yuki

if you let tan(X) = a, it's obvious that it's a quadratic such that (a-3)(a-2) = 0 so,

3. Yuki

$\tan(x) = 3,-1$

4. anonymous

would it be (a+1)(a-3) ?

5. anonymous

sorry, you did get that

6. Yuki

there are infinitely many solutions for x, but since the interval is only from 0 to 2pi, x = $\tan^{-1}(3), \tan^{-1}(3) + \pi , \tan^{-1}(-1)$

7. Yuki

you need a calculator for arctan(3) but arctan(-1) is a value we know which is -45 degrees. in terms of radians in the 3rd quadrant, it would be $7\pi \over 4$

8. Yuki

got it ? :)

9. anonymous

so is the answer 7pi/4 and that's it?

10. Yuki

no, arctan(3) and arctan(3)+pi are also answers

11. Yuki

it's just a number that we cannot say exactly what it is

12. anonymous

oh, because my answer has to be in [0,2pi) form.....no decimals

13. Yuki

Iris, the exact answer IS arctan(3), arctan(3) + pi, 7pi/4