The general solution of the equation \[\dfrac{(1-sinx+...+{{(-1)}^{n}}sin{{x}^{n}}+..)}{(1+sinx+...+sin{{x}^{n}}+..)}=\dfrac{1-cos2x}{1+cos2x}\] $x\ne (2n+1)\dfrac{\pi }{2},n\in I$

Hint: We always had a motive of decreasing the size of the equation in trigonometry as much as we can do. Same we have to do here. Moreover we had to find out the general solution for this condition will always be followed. So we will decrease the size of the L.H.S side of the equation as much as we can and then solve it by the R.H.S side to find the value of x. In order to find out for what values of x the condition will be valid.

Complete step by step answer:
So moving further to solve the equation. First going with the L.H.S side i.e. \[\dfrac{(1-sinx+...+{{(-1)}^{n}}sin{{x}^{n}}+..)}{(1+sinx+...+sin{{x}^{n}}+..)}\] try to short it as much as we can.
On elobrating the L.H.S side , we will get \[\dfrac{(1-sinx+sin{{x}^{2}}-sin{{x}^{3}}+sin{{x}^{4}}...+{{(-1)}^{n}}sin{{x}^{n}}+..)}{(1+sinx+sin{{x}^{2}}+sin{{x}^{3}}+sin{{x}^{4}}...+sin{{x}^{n}}+..)}\]and this will continue till last.
Now from the numerator part separate the negative and positive part i.e. $\sin x$ with odd and even power respectively. So, now it will form \[\begin{align}
  & \dfrac{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.....)+(-sinx-sin{{x}^{3}}-sin{{x}^{5}}...+sin{{x}^{n}}+..)}{(1+sinx+sin{{x}^{2}}+sin{{x}^{3}}+sin{{x}^{4}}...+sin{{x}^{n}}+..)} \\
 & \dfrac{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.....)-(sinx+sin{{x}^{3}}+sin{{x}^{5}}...+sin{{x}^{n}}+..)}{(1+sinx+sin{{x}^{2}}+sin{{x}^{3}}+sin{{x}^{4}}...+sin{{x}^{n}}+..)} \\
\end{align}\]
Now making the denominator part also the same, i.e. even power of $\sin x$at one side and odd power at another side. So, it will be of form;
\[\begin{align}
  & \dfrac{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.....)-(sinx+sin{{x}^{3}}+sin{{x}^{5}}...+sin{{x}^{n}}+..)}{(1+sinx+sin{{x}^{2}}+sin{{x}^{3}}+sin{{x}^{4}}...+sin{{x}^{n}}+..)} \\
 & \dfrac{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.....)-(sinx+sin{{x}^{3}}+sin{{x}^{5}}...+sin{{x}^{n}}+..)}{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}})+(sinx+sin{{x}^{3}}+sin{{x}^{5}}...+sin{{x}^{n}}+..)} \\
 & \dfrac{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.....)-sinx(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.......)}{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}})+sinx(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}......)} \\
\end{align}\]
Out of which $\sin x$is taken common from the sequence of odd powers and we got the remaining one same term as we had with even power. Now take this same term common from both numerator and denominator, so it will be;
\[\begin{align}
  & \dfrac{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.....)-sinx(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.......)}{(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}})+sinx(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}......)} \\
 & \dfrac{(1-sinx)(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}.......)}{(1+sinx)(1+sin{{x}^{2}}+sin{{x}^{4}}+sin{{x}^{6}}......)} \\
 & \dfrac{(1-sinx)}{(1+sinx)} \\
\end{align}\]
As these same term will get cancel and we will get \[\dfrac{(1-sinx)}{(1+sinx)}\]and now which is equal to \[\dfrac{1-cos2x}{1+cos2x}\]; so it will be
\[\dfrac{(1-sinx)}{(1+sinx)}=\dfrac{1-cos2x}{1+cos2x}\]
By applying Componendo and dividendo, we will get
\[\begin{align}
  & \dfrac{(1-sinx)}{(1+sinx)}=\dfrac{1-cos2x}{1+cos2x} \\
 & \dfrac{1}{sinx}=\dfrac{1}{cos2x} \\
 & cos2x=sinx \\
\end{align}\]
As we know that \[cos2x=1-2sin{{x}^{2}}\]
So;\[\begin{align}
  & cos2x=sinx \\
 & 1-2sin{{x}^{2}}=sinx \\
\end{align}\]
Let, \[sinx=p\]
So, now the equation will be;
\[\begin{align}
  & 1-2sin{{x}^{2}}=sinx \\
 & 1-2{{p}^{2}}=p \\
 & 2{{p}^{2}}+p-1=0 \\
\end{align}\]
Now solving this quadratic equation in terms of p by middle term splitting, we will get;
\[\begin{align}
  & 2{{p}^{2}}+p-1=0 \\
 & {{p}^{2}}+\dfrac{p}{2}-\dfrac{1}{2}=0 \\
 & {{p}^{2}}+\left( 1-\dfrac{1}{2} \right)p+\left( 1\times \dfrac{-1}{2} \right)=0 \\
 & {{p}^{2}}+p+\dfrac{-1}{2}p-\dfrac{1}{2}=0 \\
 & p(p+1)-\dfrac{1}{2}(p+1)=0 \\
 & (p+1)\left( p-\dfrac{1}{2} \right)=0 \\
\end{align}\]
So, ‘p’ will be equal to $-1$ or $\dfrac{1}{2}$, and as we had assumed \[sinx=p\]
So, \[sinx=-1\]\[\] or \[sinx=\dfrac{1}{2}\]
Now for general solution of\[sinx=\sin y\] we know that \[x\text{ }=\text{ }n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}y\] , so now comparing above result for general solution, we will get;
\[\begin{align}
  & sinx=-1 \\
 & sinx=sin\left( \dfrac{3\pi }{2} \right) \\
 & x=\text{ }n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}y \\
 & so, \\
 & x\text{ }=\text{ }n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}\dfrac{3\pi }{2} \\
\end{align}\]
And similarly for\[sinx=\dfrac{1}{2}\], i.e.
\[\begin{align}
  & sinx=\dfrac{1}{2} \\
 & sinx=sin\left( \dfrac{\pi }{6} \right) \\
 & As;x\text{ }=\text{ }n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}y \\
 & So, \\
 & x\text{ }=\text{ }n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}\dfrac{\pi }{6} \\
\end{align}\]
Where n belongs to integers.
Hence our answer is \[n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}\]and \[n\pi \text{ }+\text{ }{{\left( -1 \right)}^{n}}\dfrac{3\pi }{2}\]

Note: For it you had to remember the formula for the general solution of \[sinx=\sin y\]and should know the middle term splitting, which is the part of quadratic equations. Rest we had to solve as we use to solve general trigonometric expressions.