Expand.
\[2\sin{x}\cos{x}+1-1=2sin\cos{x}\]
Simplify \(2\sin{x}\cos{x}+1-1\) to \(2\sin{x}\cos{x}\).
\[2\sin{x}\cos{x}=2sin\cos{x}\]
Cancel \(2\) on both sides.
\[\sin{x}\cos{x}=sin\cos{x}\]
Move all terms to one side.
\[\sin{x}\cos{x}-sin\cos{x}=0\]
Factor out the common term \(\cos{x}\).
\[\cos{x}(\sin{x}-sin)=0\]
Solve for \(x\).
Ask: When will \(\cos{x}(\sin{x}-sin)\) equal zero?
When \(\cos{x}=0\) or \(\sin{x}-sin=0\)
Solve each of the 2 equations above.
\[\begin{aligned}&x=0\\&x=2\pi n+\sin^{-1}{(sin)},n \in Z\\&x=2\pi n+\pi -\sin^{-1}{(sin)},n \in Z\end{aligned}\]
\[\begin{aligned}&x=0\\&x=2\pi n+\sin^{-1}{(sin)},n \in Z\\&x=2\pi n+\pi -\sin^{-1}{(sin)},n \in Z\end{aligned}\]
x=0,sequence(2*PI*n+arcsin(sin),in(n,Z)),sequence(2*PI*n+PI-arcsin(sin),in(n,Z))
