Is There an Analytical Method for Solving the Given Equation?

I recently encountered this problem on a problem set:

Given that $$\log_2(2 + \sqrt{3}) < \log(3\pi)$$ find the number of roots of the equation $$4\cos(e^x) = 2^x + 2^{-x}$$

I consulted my teacher for the solution and he checked the solution he had and told me that it was not possible to solve this manually as the solution he had was via graphing the functions and it was not possible without a graphing calculator or an app.

While using a graphing calculator the initial inequality is not used but it must have been given for a reason.

Graphically 4 solutions are obtained.

I want to know if there is really no way to solve this question using only calculations which are doable by hand.

• Thanks a lot for editing this
– user671231
Commented Jul 16, 2019 at 21:09
• What do you mean in the title about solving the inequality? There is no variable to solve for. It is either true or false. In this case it is true. Commented Jul 17, 2019 at 2:35
• @RossMillikan no I had a different title it was edited by someone I'll re-edit it
– user671231
Commented Jul 17, 2019 at 9:51

Call the LHS and RHS of the equation

$$L(x) = 4\cos(e^x)$$

and

$$R(x) = 2^x+2^{-x}$$

respectively.

On the interval $$[0,\ln{\pi}]$$, $$L(x)$$ strictly decreases from $$4\cos{1}\approx2.16$$ to $$-4$$. On the same interval, $$R(x)$$ strictly increases from $$2$$ to $$2^{\ln{\pi}}+2^{-\ln{\pi}}\approx2.66$$. We then deduce that there is exactly one root in $$[0,\ln{\pi}]$$.

On the interval $$[\ln{\pi},\ln{\frac{3}{2}\pi}]$$, $$L(x)\le0$$ while $$R(x)\gt0$$ so there are zero roots in this interval.

On the interval $$[\ln{\frac{3}{2}\pi},\ln{\frac{5}{2}\pi}]$$, $$L(x)$$ is concave down ($$L''\lt0$$). On the same interval, $$R(x)$$ is concave up ($$R''\gt0$$). So there are either 0 or 2 roots in this interval. At $$x=\ln{2\pi}$$, $$L(x)=4$$ is greater than $$R(x)=2^{\ln{2\pi}}+2^{-2\ln{\pi}}\approx3.85$$. Yet on the ends of the interval $$L(x) \lt R(x)$$. We deduce that there are exactly 2 roots in this interval.

On the interval $$[\ln{\frac{5}{2}\pi},\infty)$$ there are no more roots because $$R(x)>4$$ and $$L(x)\le4$$.

For $$x\lt0$$ as $$x\rightarrow-\infty$$, $$L(x)$$ asymptotically approaches $$4$$ with continuously decreasing (negative) slope. In the same interval $$R(x)$$ is increasing with continuously increasing (negative) slope. We deduce that there is exactly one root in this interval.

For a total of four roots.

• Thanks a lot. Can you please explain how did you confirm the number of roots in the interval [lnπ,ln2π]?
– user671231
Commented Jul 17, 2019 at 15:17
• I didn't exactly. As noted in my answer, I claimed that there is at least one root in that interval. After further thought, I think I can improve the answer and get the exact number of roots. I will update the answer. Commented Jul 17, 2019 at 17:40
• Ok got it thanks a lot looking forward for the update
– user671231
Commented Jul 17, 2019 at 17:42