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Find all $\theta \in \Bigl (-\dfrac{\pi}{2},\dfrac{\pi}{2}\Bigr)$ satisfying:

$$\sec ^{2} \theta(1-\tan \theta)(1+\tan \theta)+2^{\tan^{2}\theta}=0$$

I have tried a lot but couldn't crack this one. I could only bring it down to the following problem (Solving the following problem is equivalent to solving the above equation):

Find all $t \in \mathbb R^{+}$ satisfying $$\begin{align} t^{2}=2^{t}+1 \tag{1}\end{align}$$

Any suggestions on how to solve either of the two problems? By plotting a rough graph, I could figure out that there are two such $t$'s satisfying $(1)$, but which ones?

Thanks for the help.

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Apurv
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  • There are actually 3 solutions, according to WolframAlpha. Not sure how to prove them though. – 2012ssohn Jan 20 '14 at 13:59
  • One easy solution of (1) is $t=3$ – gammatester Jan 20 '14 at 14:11
  • @gammatester, I have already observed that.. But any ideas on how to find all the solutions?? – Apurv Jan 20 '14 at 14:13
  • If you look at the function $f(t) = t^2 -2^t -1$ there are three zero, the already recognized $t=3$ and the approximates $t\approx 3.40745$ and $t \approx -1.19825$. If should be relative easy to show that there are no other roots considering the rapid growth rate of $2^t.$ – gammatester Jan 20 '14 at 14:28
  • any ideas on how to find all the solutions?? – Apurv Jan 21 '14 at 04:10
  • Please don't use double dollar signs $$ ... $$ in titles. – Antonio Vargas Jan 22 '14 at 03:02
  • @AntonioVargas, sorry.. Any ideas on the problem?? – Apurv Jan 22 '14 at 03:05
  • From http://math.stackexchange.com/questions/646092/number-of-solutions-of-px-eax-if-p-is-a-polynomial we know that there are at most three solutions. Therefore the three given are the solutions. The root at $3.446\dots$ can be computed numerically, e.g. with http://en.wikipedia.org/wiki/Brent%27s_method I get $3.407450522185998622$ with 13 function calls: all shown digits correct, own Pascal library, starting with the interval $[3.1,5]$. – gammatester Jan 22 '14 at 08:57

1 Answers1

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The function $f(t)=t^2−2^t−1$ has three zeros. The obvious $t=3$ and two others. Since the t comes from the substitution of $\tan^2$, we can ignore the negative zero.

As to why there are three zeros, sketching a graph and knowing how the exponential function and quadratic function behave would suffice. $(t^2 - 1 = 2^t)$

As for the third zero $(t≈3.40745)$ if you are on an exam system using graphical calculators, then an approximation from the graph is acceptable. Otherwise, use iterative formula with a starting value of any value greater than $4$.

e.g Starting with $t=3.5$ and using $t=\log(t^2 -1)/\log2$, I get $3.446115936$ after 10 iterations.

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