Question

2sin^2theta-sin theta= 0 with intervals 0 less than or equal to theta less than or equal to 360

Answer 1

\(2\sin^2 \theta - \sin \theta=0 \Leftrightarrow \sin \theta(2\sin \theta - 1)=0\)

Now solve separately sinθ = 0 and 2sinθ−1=0.

Answer 2

Sin theta= 0 and sin=1/2

Answer 3

Yes, so what can theta be then?