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

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Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0sec^2 = 1 + tan^2 so if you substitute that you will get \[\tan^2(x) 2\tan(x) 3 = 0\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0if you let tan(X) = a, it's obvious that it's a quadratic such that (a3)(a2) = 0 so,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it be (a+1)(a3) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, you did get that

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0there 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
 5 years ago
Best ResponseYou've already chosen the best response.0you 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 \[7\pi \over 4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is the answer 7pi/4 and that's it?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0no, arctan(3) and arctan(3)+pi are also answers

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0it's just a number that we cannot say exactly what it is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, because my answer has to be in [0,2pi) form.....no decimals

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0Iris, the exact answer IS arctan(3), arctan(3) + pi, 7pi/4
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