anonymous
  • anonymous
Solve: sec²θ = 2tanθ+4 over the interval [0,2pi)
Mathematics
jamiebookeater
  • jamiebookeater
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yuki
  • yuki
sec^2 = 1 + tan^2 so if you substitute that you will get \[\tan^2(x) -2\tan(x) -3 = 0\]
yuki
  • yuki
if you let tan(X) = a, it's obvious that it's a quadratic such that (a-3)(a-2) = 0 so,
yuki
  • yuki
\[\tan(x) = 3,-1\]

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anonymous
  • anonymous
would it be (a+1)(a-3) ?
anonymous
  • anonymous
sorry, you did get that
yuki
  • yuki
there are infinitely many solutions for x, but since the interval is only from 0 to 2pi, x = \[\tan^{-1}(3), \tan^{-1}(3) + \pi , \tan^{-1}(-1)\]
yuki
  • 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 \[7\pi \over 4\]
yuki
  • yuki
got it ? :)
anonymous
  • anonymous
so is the answer 7pi/4 and that's it?
yuki
  • yuki
no, arctan(3) and arctan(3)+pi are also answers
yuki
  • yuki
it's just a number that we cannot say exactly what it is
anonymous
  • anonymous
oh, because my answer has to be in [0,2pi) form.....no decimals
yuki
  • yuki
Iris, the exact answer IS arctan(3), arctan(3) + pi, 7pi/4

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