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

1. ## 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.

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

Originally Posted by Sjm2015
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 \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 \displaystyle \begin{align*} x = \sin{t} \end{align*} to give \displaystyle \displaystyle \begin{align*} \sqrt{2}\,x^2 + x - \sqrt{2} = 0 \end{align*}, solve for \displaystyle \displaystyle \begin{align*}x \end{align*}, then solve for \displaystyle \displaystyle \begin{align*} t \end{align*}.

3. ## 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.

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

Originally Posted by Sjm2015
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 \displaystyle \begin{align*} \sqrt{2} \cdot \cot{t} \end{align*} or \displaystyle \displaystyle \begin{align*} \sqrt{2 \cdot \cot{t}} \end{align*}, or something else entirely?

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

one equation, one conditional inequality ...

$\displaystyle \sec{t} = \sqrt{2}$

$\displaystyle \cot{t} > 0$

6. ## 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 $\displaystyle sec(t)= \sqrt{2}$ Then $\displaystyle cos(t)= \frac{1}{sec(t)}= \frac{1}{\sqrt{2}}$ That is true for $\displaystyle t= \frac{\pi}{4}$ and $\displaystyle t= \frac{7\pi}{4}$ but only the first is less than 2 and has cot(t)> 0.