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?
The problem might also be approached as a search for the intersections of the function curves $ \ y \ = \ e^{2x} \ + \ 3e^x \ + \ 1 \ = \ 0 \ $ and $ \ y \ = \ 5 e^x \cos x \ \ . $ It is then helpful to consider the properties of these functions.
The former is the sum of a constant term and two "exponential-growth" functions, so it is monotonically increasing from its asymptotic value of $ \ 1 \ \ . $ Thus, this function value is very close to $ \ 1 \ $ until around $ \ x = -2 \ \ , $ where it has increased to $ \ 1 + 3e^{-2} + e^{-4} \ \approx \ 1.4 \ \ . $ The curve has a $ \ y-$intercept of $ \ (0 \ , \ 5 ) \ , $ has risen to $ \ y \ \approx \ 16.5 \ $ at $ \ x \ = \ 1 \ \ $ and continues to increase ever after.
The latter is the product of an "exponential-growth" function and a sinusoidal function, so its values fall within the "envelope" $ \ -5e^x \ \le \ y \ \le \ 5e^x \ \ . $ The function is then very close to zero until $ \ x \ $ becomes a fairly small negative number. The upper limit of the envelope only reaches $ \ 1 \ $ at $ \ x \ = \ \ln \frac15 \ \approx \ -1.60 \ \ , $ but the value of the function is close to zero there, with the curve having an $ \ x-$intercept at $ \ x \ = \ -\frac{\pi}{2} \ \approx \ -1.57 \ \ . $ The envelope's upper limit is $ \ y = 5 \ \ $ at the $ \ y-$axis, where the curve rises within it to its $ \ y-$intercept at $ \ (0 \ , \ 5) \ \ . $ Thus, the two function curves have an intersection (at last) at that point.
Beyond the $ \ y-$axis, $ \ e^{2x} \ + \ 3e^x \ + \ 1 \ $ continues to increase indefinitely, while $ \ 5 e^x \cos x \ $ immediately "falls behind" and begins to "drop" toward its next $ \ x-$intercept at $ \ x \ = \ \frac{\pi}{2} \ \ . $ (This is a place where we would need to resort to calculus to show that the rate of increase of this second function is less than that of the first at $ \ x = 0 \ \ . ) $ While the "size" of its oscillations continue to grow without limit, this function can now never "catch up" with the monotonically increasing function. (We observe that for $ \ x \ > \ \ln 5 \ \approx \ +1.60 \ \ , $ $$ e^{2x} \ + \ 3e^x \ + \ 1 \ > \ e^{2x} \ = \ e^x · e^x \ > \ 5 · e^x \ \ . ) $$
Thus, there is only the single intersection that you have found.