[tex]-1\le\sin\theta\le1\implies(\csc\theta\le-1)\cup(\csc\theta\ge1)[/tex]
which means [tex]\csc\theta=0[/tex] has no solution, and we can omit that factor.
This leaves us with
[tex]2\cos\theta-1=0\implies\cos\theta=\dfrac12[/tex]
which occurs for [tex]\theta=\pm\dfrac\pi3+2n\pi[/tex].
