Finding number of real solution of $4^x-2^{2+x}\cos(e^x)+1=0$

Question

Finding number of real solution of $4^x-2^{2+x}\cos(e^x)+1=0$

Try:$$(2^x)^2-4\cos(e^x)(2^x)+1=0$$

$$2^x=\frac{4\cos(e^x)\pm \sqrt{16\cos^2(e^x)-4}}{2}$$

$$2^x=2\cos(e^x)\pm \sqrt{4\cos^2(e^x)-1}>0$$

Could some help me to solve it. Thanks in advance

Answer 1

Rewrite our equation in the following form: $f(x)=g(x)$, where $$f(x)=\frac{1}{4}\left(2^x+2^{-x}\right)$$ and $$g(x)=\cos{e^x}.$$

We need $|f(x)|\leq1,$ which gives $$\log_2(2-\sqrt3)\le x\leq\log_2(2+\sqrt3).$$

But for these values of $x$ we see that $g$ decreases and $g$ is a concave function for $x\leq0$.

By the way $f$ is a convex function, decreases on $\left[\log_2(2-\sqrt3),0\right]$ and increases on $\left[0, \log_2(2+\sqrt3)\right]$

Thus, our equation has two roots maximum: one on $\left(\log_2(2-\sqrt3),0\right)$ and one on $\left(0,\log_2(2+\sqrt3)\right)$.

Now, since $f$ and $g$ they are continuous functions, $$f\left(\log_2(2-\sqrt3)\right)>g\left(\log_2(2-\sqrt3)\right),$$ $$f(0)<g(0)$$ and $$f\left(\log_2(2+\sqrt3)\right)>g\left(\log_2(2+\sqrt3)\right),$$ we see that our equation has two roots exactly.