You seem to have some serious problems with the algebra involved. Part of the problem is failure to use necessary parentheses: the result of applying the quotient rule is
$$\frac{2\sin x-(2x+1)\cos x}{\sin^2x}\;,$$
where the parentheses around $2x+1$ are absolutely necessary. If you choose to multiply out the numerator, you should get
$$\frac{2\sin x-2x\cos x-\cos x}{\sin^2x}\;.$$
Alternatively, you can split it into two fractions:
$$\begin{align*}
\frac{2\sin x-(2x+1)\cos x}{\sin^2x}&=\frac{2\sin x}{\sin^2x}-\frac{(2x+1)\cos x}{\sin^2x}\\\\
&=2\csc x-(2x+1)\cot x\csc x\\\\
&=\csc x\Big(2-(2x+1)\cot x\Big)\;.
\end{align*}$$