$(2\sin x-1)(2\cos2x+2\sin x+3)+1=4\sin^2x$
$\to (2\sin x-1)(2\cos 2x+2\sin x+3)=4\sin^2x-1=(2\sin x-1)(2\sin x+1)$
$\to (2\sin x-1)(2\cos2x+2\sin x+3-2\sin x-1)=0$
$\to (2\sin x-1)(2\cos2x+2)=0$
$\to \left[ \begin{array}{l}\sin x=\dfrac{1}{2}\\\cos2x=-1 \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi \\ x=\dfrac{5\pi}{6}+k2\pi \\ 2x=\pi+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi \\ x=\dfrac{5\pi}{6}+k2\pi \\x=\dfrac{\pi}{2}+k\pi \end{array} \right.$
