Official solution
2cos2\[\theta \] + 3sin\[\theta \] = 0
\[ \Rightarrow \] 2(1 - sin2\[\theta \]) + 3sin\[\theta \] = 0
\[ \Rightarrow \] 2sin2\[\theta \] - 3sin\[\theta \] - 2 = 0
\[ \Rightarrow \] 2sin2\[\theta \] - 4sin\[\theta \] + sin\[\theta \] - 2 = 0
\[ \Rightarrow \] (2sin\[\theta \] + 1)(sin\[\theta \] - 2) = 0
\[ \therefore \] sin\[\theta \] = 2 (Not possible)
sin\[\theta \] = \[ - {1 \over 2}\]
\[ \therefore \] \[\theta \] = n\[\pi \] + (-1)n\[\left( { - {\pi \over 6}} \right)\]
When n = 0, \[\theta \] = \[{ - {\pi \over 6}}\]
When n = 1, \[\theta \] = \[{{7\pi } \over 6}\]
When n = -1, \[\theta \] = \[{-{5\pi } \over 6}\]
When n = 2, \[\theta \] = \[{{11\pi } \over 6}\]
\[ \therefore \] Sum of all solutions in [--2\[\pi \], 2\[\pi \]] is
= \[ - {\pi \over 6} + {{7\pi } \over 6} - {{5\pi } \over 6} + {{11\pi } \over 6}\] = 2\[\pi \]