How do you solve the equation $3x^2+2\sin x=2$? This seems like a simple equation to solve, but I can't find a way to do it and wolfram alpha can't give me an exact answer.

This is a transcendental equation; so, you will need numerical methods and then good starting guesses.
Consider that you look for the zero's of function $f(x)=3x^2+2\sin x-2$.
For example, using Taylor series built around $x=0$
$$f(x)=-2+2 x+3 x^2-\frac{x^3}{3}+O\left(x^5\right)$$ Ignoring the term in $x^3$, solving the quadratic would give as estimates
$$x_1=-\frac{\sqrt{7}+1}{3} \qquad \qquad x_2=\frac{\sqrt{7}-1}{3}$$ We could solve the cubic but this is not very pleasant.
But what we can do is to take quite many terms in the series expansion and use series reversion. Looking at the values of $x_1$ and $x_2$, reasonable base points could be $-\frac \pi 3$ and $\frac  \pi 6$. This would give as approximation
$$x_1=-\frac{\pi }{3}+t+\frac{\left(6+\sqrt{3}\right) }{2 (2 \pi
   -1)}t^2+\frac{\left(59+18 \sqrt{3}-\pi \right) }{3 (2 \pi
   -1)^2}t^3+O\left(t^4\right)$$ where $t=\frac{\pi ^2-3 \left(2+\sqrt{3}\right)}{3(2 \pi -1)}$. Numerically, this gives $x_1\approx -1.12638$ while Newton method would give as solution $x_1= -1.12630$.
Doing the same for the second root
$$x_2=\frac{\pi }{6}+t-\frac{5 }{2 \left(\sqrt{3}+\pi \right)}t^2+\frac{\left(78+\sqrt{3}
   \pi \right) }{6 \left(\sqrt{3}+\pi \right)^2}t^3+O\left(t^4\right)$$ where $t=\frac{1-\frac{\pi ^2}{12}}{\sqrt{3}+\pi }$.  Numerically, this gives $x_2\approx 0.559374$ while Newton method would give as solution $x_2= 0.559372$.
For sure, at the price of more messy terms, we could be even much closer to the solution.
