Intersections of $y=\cos(2x)$ and $y=e^{3x}$ such that $x \in [-\pi, \frac{\pi}{2}]$ I am having a hard time finding the intersections of $y = \cos(2x)$ and $y = e^{3x}$ such that $x \in [-\pi, \frac{\pi}{2}]$ through solving manually. But through graphing, I know that the intersections are (0,1) and (-0.7292,0.1122). I need those to find the area bounded by the two curves. Can someone help me find it through solving?

$$\cos 2x = e^{3x}$$
it can be solved finding the zeros of the function
$$f(x)=\cos 2x -e^{3x}$$
which is not an algebraic equation.
We can find approximate roots with the Newton's method
looking at the figure below, one root is about $x_0=-0.8$
So we define for $n\ge 0$
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
and reiterate the process until the error is less than $10^{-6}$ if we want $6$ exact decimals.
$$
\begin{array}{c|r|r}
 n & x_n & error\\
\hline
 0 & -0.8 &  \\
 1 & -0.730563 & 0.0694371 \\
 2 & -0.729185 & 0.00137768 \\
 3 & -0.729184 & 8.3\times 10^{-7}\\
\end{array}
$$
In a similar way we can find the second root, which is about $z_0=-2.4$
$$
\begin{array}{c|r|r}
 n & x_n & error\\
\hline
 0 & -2.4 &  \\
 1 & -2.35651 & 0.0434943 \\
 2 & -2.35662 & 0.000113975 \\
 3 & -2.3566196667 & 3.4\times 10^{-11}\\
\end{array}
$$
The two solutions in the given interval are:
$$\alpha_1=-2.3566197,\alpha_2=-0.729184$$


