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anonymous

  • 5 years ago

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

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  1. Yuki
    • 5 years ago
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    sec^2 = 1 + tan^2 so if you substitute that you will get \[\tan^2(x) -2\tan(x) -3 = 0\]

  2. Yuki
    • 5 years ago
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    if you let tan(X) = a, it's obvious that it's a quadratic such that (a-3)(a-2) = 0 so,

  3. Yuki
    • 5 years ago
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    \[\tan(x) = 3,-1\]

  4. anonymous
    • 5 years ago
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    would it be (a+1)(a-3) ?

  5. anonymous
    • 5 years ago
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    sorry, you did get that

  6. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    got it ? :)

  9. anonymous
    • 5 years ago
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    so is the answer 7pi/4 and that's it?

  10. Yuki
    • 5 years ago
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    no, arctan(3) and arctan(3)+pi are also answers

  11. Yuki
    • 5 years ago
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    it's just a number that we cannot say exactly what it is

  12. anonymous
    • 5 years ago
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    oh, because my answer has to be in [0,2pi) form.....no decimals

  13. Yuki
    • 5 years ago
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    Iris, the exact answer IS arctan(3), arctan(3) + pi, 7pi/4

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