The question is to find the number of solutions of the equation $e^{\sin x+\cos x}=x^2+x+1$
To which I simplified to $\sin x+\cos x=\ln(x^2+x+1)$
I can graph $\sin x+\cos x$ quite easily. But I struggle in graphing $\ln(x^2+x+1)$
If a method could be given to graph such a function that would be highly appreciated.
The red graph is that of $\ln(x^2+x+1)$, and the blue graph (easy to draw) is for $\sin x + \cos x$. How to draw $f(x) = \ln(x^2+x+1)$? Let's see.
I would start by noticing $f(0)=0$. Then I would find $$f'(x) = \frac{2x+1}{x^2+x+1}$$
Note that $x^2+x+1 = (x+0.5)^2 + 0.75 > 0$ for all $x\in\mathbb{R}$. So the sign of $f'(x)$ is determined by the sign of $2x+1$. $f'(x)>0$ for $x>-0.5$, $f'(x) = 0$ for $x=-0.5$ and negative otherwise. We deduce that $x=-0.5$ is a point of local minima for $f(x)$ and $f(x)$ strictly increases for $x>-0.5$ (and strictly decreases for $x<-0.5$). All that remains is to find $f(-0.5)$ which I'm sure you can do.
In the end, you get the graph below and exactly two solutions for the equation.
P.S.
It is worth noting that $g(x) = \ln x$ is an increasing function, so it would actually be enough to study the monotonicity of $h(x) = x^2+x+1$ to draw the graph of $f(x) = g(h(x))$. I hope you can see why.