I have a literal equation that needs to be solved for $\theta$:
$$mg \sin(\theta) = \mu mg \cos(\theta)\left({ M+m \over m}\right) $$
I have a literal equation that needs to be solved for $\theta$:
$$mg \sin(\theta) = \mu mg \cos(\theta)\left({ M+m \over m}\right) $$
A simple rearrangement shows that $$\tan\theta = \mu\left(1+\frac{M}{m}\right)$$ to obtain numerical values for the equation you'll need the ratio of the masses and the coefficient of friction, in which case you can simply take the arctangent.