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Math Help - sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

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    sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

    sec(t) = √2 cot(t) >0. Find all values of t in the interval [0,2).

    I've been looking at examples of similar problems like this but I'm just not getting it. Would sec(t)=√2 be flipped as cos(t)=1/√2? Any help would be great. Lost as to what next step would be.
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    Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

    Quote Originally Posted by Sjm2015 View Post
    sec(t) = √2 cot(t) >0. Find all values of t in the interval [0,2).

    I've been looking at examples of similar problems like this but I'm just not getting it. Would sec(t)=√2 be flipped as cos(t)=1/√2? Any help would be great. Lost as to what next step would be.
    \displaystyle \begin{align*} \sec{t} &= \sqrt{2}\cot{t} \\ \frac{1}{\cos{t}} &= \frac{\sqrt{2}\cos{t}}{\sin{t}} \\ \sin{t} &= \sqrt{2}\cos^2{t} \\ \sin{t} &= \sqrt{2}\left( 1 - \sin^2{t}\right) \\ \sin{t} &= \sqrt{2} - \sqrt{2}\sin^2{t} \\ \sqrt{2}\sin^2{t} + \sin{t} - \sqrt{2} &= 0 \end{align*}

    Let \displaystyle \begin{align*} x = \sin{t} \end{align*} to give \displaystyle \begin{align*} \sqrt{2}\,x^2 + x - \sqrt{2} = 0 \end{align*}, solve for \displaystyle \begin{align*}x  \end{align*}, then solve for \displaystyle \begin{align*} t \end{align*}.
    Last edited by Prove It; August 21st 2012 at 08:38 PM.
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    Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

    Oh shoot, there was suppose to be a period seperating the two. sec(t) =√2. cot(t) >0. But thank you, that helped me with another problem lol.
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    Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

    Quote Originally Posted by Sjm2015 View Post
    Oh shoot, there was suppose to be a period seperating the two. sec(t) =√2. cot(t) >0. But thank you, that helped me with another problem lol.
    I don't see how that makes any difference to what I wrote. Is the RHS \displaystyle \begin{align*} \sqrt{2} \cdot \cot{t} \end{align*} or \displaystyle \begin{align*} \sqrt{2 \cdot \cot{t}} \end{align*}, or something else entirely?
    Last edited by Prove It; August 21st 2012 at 09:25 PM.
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    Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

    one equation, one conditional inequality ...

    \sec{t} = \sqrt{2}

    \cot{t} > 0
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    Re: sec(t) =√2 cot(t) >0. Find all values of t in the interval [0,2)

    If it is, in fact, as skeeter suggests, to find t such that cot(t)> 0 and sec(t)= \sqrt{2} Then cos(t)= \frac{1}{sec(t)}= \frac{1}{\sqrt{2}} That is true for t= \frac{\pi}{4} and t= \frac{7\pi}{4} but only the first is less than 2 and has cot(t)> 0.
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