Question

Find all solutions of sin2x = ...

Answer 1

\[\sin2x=\frac{ -\sqrt{3} }{ 2 }\]

Answer 2

\[2x=\frac{ -\pi }{ 3 }\]
\[x=-\frac{ \pi }{ 6 }\]

Answer 3

If\[sin \theta=sin \alpha\] then \[\theta=n \pi +(-1)^n\alpha\]wher n=0,1,2...

Answer 4

\[sin2x=sin \frac{-\pi}{3}\]\[2x=n \pi+(-1)^n \frac{-\pi}{3}\]From here can u determine all the slution