Why can't equations with unknown inside and outside of a function be solved in a standard way? For example the equation 
$$
n2^n = 8
$$
Is true for $n=2$ which can be guessed, but more complicated examples would require some sort of approach. Also with trigonometric functions, 
$$
x\sin(x) + B\cos(x) = A
$$
I read that solutions of these kinds of equations can not be expressed in closed form, why is this the case? 
 A: It depends on what you mean by "closed form". If you allow the natural logarithm (the inverse of $y = e^x$) you can solve equations like $5^x = 7$ by writing $x = \dfrac{\ln 7}{\ln 5}$. 
In the case of expressions like $n2^n = 8$ there is a function called the Lambert $W$-function which is defined as the inverse of $y = xe^x$, $x \ge -1$. That is $x = W(y)$ is equivalent to $y = xe^x$. Then $$8 = n2^n = n e^{\ln 2} = \frac{n \ln 2 e^{n \ln 2}}{\ln 2}$$
so that $n \ln 2 e^{n \ln 2} = 8 \ln 2$, implying $n\ln 2 = W(8 \ln 2)$ and thus $n = \dfrac{W(8 \ln 2)}{\ln 2}$.
This "solves" $8 = n2^n$ in much the same way that $x=(\ln 7)/(\ln 5)$ solves $5^x = 7$.
A: First off, you need to realize that solving equations is the process of determining the inverse function.  If I have $n2^n=8$, what I'm really doing is starting with the function $f(n)=n2^n$, and that statement $f(n)=8$ means I need to find $f^{-1}(8)$. When I can find a functional expression for $f^{-1}$, then I have a closed form solution, i.e. a function that I can write down that gives me the inverse of $f$ across all of $f$'s domain.
The reason finding closed form solutions is so hard is that the process of functional inversion doesn't distribute over simple algebra.
For example, look at derivation. Derivation takes a function and returns you a different function, just like inversion does, and it behaves well with arithmetic. Using $d$ to represent the derivation operator, $d(f+g) = df + dg$ and $d(f\times g) = f\times dg + df\times g$. This works no matter how complex $f$ and $g$ are.
However, inversion doesn't play well with others. Using $I$ to represent the inversion operator, there's no general formula for $I(f+g)$ or $I(f \times g)$. That's why things get hairy even before complex functions come into play. For instance, while there are simple algebraic solutions for low order polynomials (e.g. the quadratic equation), there are entire classes of specialized functions created just to express closed form solutions for higher order polynomials.   
As an exercise: With $f(x)=2x$ and $g(x)=3x$, try finding a simple formula for the inverse of $f$ + $g$ written in terms of $f, g, f^{-1},$ and $g^{-1}$. Now try that formula using $f(x)=x^2$ and $g(x)=x^2$.
