Solve:
sec²θ = 2tanθ+4 over the interval [0,2pi)

Question

yuki · Answer

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

yuki · Answer

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

yuki · Answer

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

anonymous · Answer

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

anonymous · Answer

sorry, you did get that

yuki · Answer

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

yuki · Answer

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 

$$7\pi \over 4$$

yuki · Answer

got it ? :)

anonymous · Answer

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

yuki · Answer

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

yuki · Answer

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

anonymous · Answer

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

yuki · Answer

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