Question

(sin2x)^2-4sin2x-1=0
Will add all the steps I took

Answer 1

\[(\sin2x)^{2}-4\sin2x=1\]
\[(\sin2x)^{2}-4\sin2x=1\] +4 +4
\[\sqrt{(\sin2x)^{2}-\sin2x}==\sqrt{5}\]
\[(\sin2x)-2=+-\sqrt{5}\] +2 +2
\[(\sin2x)=2+-\sqrt{5}\]
\[2x=(sin)^{-1}(2+-\sqrt{5})\]
Now wut?..