# Solve the equation : $e^{2x}+e^{x}\left(3-5\cos x\right)+1=0$

Solve the equation

$$e^{2x}+e^{x}\left(3-5\cos x\right)+1=0$$

I by hit and trial found a solution $$x=0$$.

Clearly this is a quadratic in $$e^x$$, so first I made discriminant positive which gives

$$\cos x\leq\frac{1}{5}$$ and $$\cos x =1$$

Now $$\displaystyle e^x=\frac{3-5\cos x\pm\sqrt{5(5\cos x-1)(\cos x-1)}}{2}$$

Now since $$e^x>0$$ for all $$x\in R$$, therefore we need to do $${3-5\cos x\pm\sqrt{5(5\cos x-1)(\cos x-1)}}>0$$

which I am not able to solve.

Is this the correct way to proceed? If so then how can I proceed?

• Hint: Set $f(x)=e^{2x}+e^x(3-5\cos(x))+1$. Then $f(x)\geq e^{2x}-2e^x+1>0$ for all $x\neq 0$.
– Surb
Nov 25 at 8:08
• Consider substituting $e^x = a$, and you will get a partial quadratic. Since $e^x > 0$, what does this tell you about the functions roots? Nov 25 at 8:13
• @UnexpectedConfusion but then what about the x in cosx term Nov 25 at 8:24
• @LalitTolani $x = 0$ is a root thats why his statement is true ... and $\cos(x)$ term stays $\cos(x)$, all my point is stating that the root is minimum due to the partial quadratic nature of the function Nov 25 at 8:27
• @LalitTolani Yes, precisely. However, this inequality would lead you to the solution even without having "guessed" that $x=0$ is a solution. Knowing that $f(x)\ge (e^x-1)^2$ it is natural to check what happens to $f(x)$ when $e^x-1 =0$. Nov 25 at 8:38