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.

  • $\begingroup$ There are actually 3 solutions, according to WolframAlpha. Not sure how to prove them though. $\endgroup$ – 2012ssohn Jan 20 '14 at 13:59
  • $\begingroup$ One easy solution of (1) is $t=3$ $\endgroup$ – gammatester Jan 20 '14 at 14:11
  • $\begingroup$ @gammatester, I have already observed that.. But any ideas on how to find all the solutions?? $\endgroup$ – Apurv Jan 20 '14 at 14:13
  • $\begingroup$ 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 $2^t.$ $\endgroup$ – gammatester Jan 20 '14 at 14:28
  • $\begingroup$ any ideas on how to find all the solutions?? $\endgroup$ – Apurv Jan 21 '14 at 4:10

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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