sin²θ - 3sinθ + 2 = 0, will be true if ?

Solution (By Examveda Team)

$$\eqalign{ & {\sin ^2}\theta - 3\sin \theta + 2 = 0 \cr & \Rightarrow {\sin ^2}\theta - 2{\text{sin }}\theta - \sin \theta + 2 = 0 \cr & \Rightarrow \sin \theta \left( {\sin \theta - 2} \right) - 1\left( {\sin \theta - 2} \right) = 0 \cr & \Rightarrow \left( {\sin \theta - 1} \right)\left( {\sin \theta - 2} \right) = 0 \cr & \left[ {\because \sin \theta \ne 2} \right]{\text{Put value of }} \cr & \Rightarrow \sin \theta = 1 \cr & \Rightarrow {\text{sin }}\theta = \sin {90^ \circ } \cr & \Rightarrow \theta = {90^ \circ } \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }}\theta = {90^ \circ } \cr & \left[ {{\text{Take help from the options}}} \right] \cr & \Rightarrow {\sin ^2}\theta - 3\sin {\text{ }}\theta + 2 = 0 \cr & \Rightarrow {\sin ^2}{90^ \circ } - 3{\text{sin }}{90^ \circ } + 2 = 0 \cr & \Rightarrow 1 - 3 \times 1 + 2 = 0 \cr & \left[ {\sin {{90}^ \circ } = 1} \right] \cr & \Rightarrow 0 = 0\left[ {{\text{matched}}} \right] \cr & {\text{So, this is answer}}{\text{.}} \cr} $$