sin 2θ = cos 3θ, then find θ = ?

1
\(\left( {4n + 1} \right)\frac{\pi }{{10}}\)
2
\(\left( {4n - 1} \right)\frac{\pi }{{10}}\)
3
\(\left( {4n + 1} \right)\frac{\pi }{5}\)
4
\(\left( {4n - 1} \right)\frac{\pi }{5}\)

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