Solve for x (complex solution)
\left\{\begin{matrix}x=-\frac{2ye^{i-2it}}{t\left(i-ie^{2\left(-2it+i\right)}\right)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{2}+\frac{1}{2}\text{ and }t\neq 0\\x\in \mathrm{C}\text{, }&\left(\exists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{2}+\frac{1}{2}\text{ or }t=0\right)\text{ and }y=0\end{matrix}\right.
Solve for x
\left\{\begin{matrix}x=\frac{y}{t\sin(2t-1)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{2}+\frac{1}{2}\text{ and }t\neq 0\\x\in \mathrm{R}\text{, }&y=0\text{ and }\left(\exists n_{1}\in \mathrm{Z}\text{ : }t=\frac{\pi n_{1}}{2}+\frac{1}{2}\text{ or }t=0\right)\end{matrix}\right.
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tx\sin(2t-1)=y
Swap sides so that all variable terms are on the left hand side.
t\sin(2t-1)x=y
The equation is in standard form.
\frac{t\sin(2t-1)x}{t\sin(2t-1)}=\frac{y}{t\sin(2t-1)}
Divide both sides by t\sin(2t-1).
x=\frac{y}{t\sin(2t-1)}
Dividing by t\sin(2t-1) undoes the multiplication by t\sin(2t-1).
tx\sin(2t-1)=y
Swap sides so that all variable terms are on the left hand side.
t\sin(2t-1)x=y
The equation is in standard form.
\frac{t\sin(2t-1)x}{t\sin(2t-1)}=\frac{y}{t\sin(2t-1)}
Divide both sides by t\sin(2t-1).
x=\frac{y}{t\sin(2t-1)}
Dividing by t\sin(2t-1) undoes the multiplication by t\sin(2t-1).
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