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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?

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    $\begingroup$ 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 $x\neq 0$. $\endgroup$
    – Surb
    Nov 25 at 8:08
  • $\begingroup$ 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? $\endgroup$ Nov 25 at 8:13
  • $\begingroup$ @UnexpectedConfusion but then what about the x in cosx term $\endgroup$ Nov 25 at 8:24
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    $\begingroup$ @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 $\endgroup$ Nov 25 at 8:27
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    $\begingroup$ @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 $e^x-1 =0$. $\endgroup$ Nov 25 at 8:38

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