Number of real roots of equation $e^{\cos x} - e^{-\cos x} - 4 = 0$?

Question

I don't have much knowledge of Euler's number. How do I approach this problem? I tried taking the logarithm of both sides and ended up with $$ \log_e e^{2 \cos x} = \log_e 4 $$ giving $$ 2 \cos x = \log_e 4 $$ but I don't know if that is correct, or what I should do next. How should I proceed?

Answer 1

Let $y=e^{\cos x}$.

Your equation becomes $y-\dfrac1y=4$, or $y^2-4y-1=0$.

This will yield two real solutions, $y=2\pm\sqrt5$.

Then, you have to solve $e^{\cos x}=y$, or $\cos x=\log y$, only valid if $y>0$ (the preceding equation has only one positive solution, $y=2+\sqrt5$).

The equation $\cos x=\log y$ has solutions only if $|\log y|\le1$, but you can check that $2+\sqrt5>4>e$, hence $\log (2+\sqrt5)>1$ and your equation has no real solution.

Answer 2

Let $ \cos x =u$ which is a function bounded within $\pm 1.$

This is same as $\sinh (u)=2$ which has a solution $x=1.44$ approximately (graph cuts at a single point); so there is no real solution.

Answer 3

$$e^{\cos x}-e^{-\cos x}<e^{\cos x}\leq e <4$$

$$\implies e^{\cos x}-e^{-\cos x}=4$$ has no real solution.