0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.