Giúp với 1 câu thôi Sin(2x+1)=cos(3x+2)

`sin (2x + 1) = cos (3x + 2)`

`<=> sin (2x + 1) = sin (π/2 – 3x – 2)`

`<=>`  \(\left[ \begin{array}{l}2x + 1 = \dfrac{π}{2} – 3x – 2 + k2π\\2x + 1 = π – \dfrac{π}{2} + 3x + 2 + k2π\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}5x = \dfrac{π}{2} – 3 + k2π\\-x = \dfrac{π}{2} + 1 + k2π\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{10} – \dfrac{3}{5} + k\dfrac{2π}{5}\\x = -\dfrac{π}{2} – 1 + k2π\end{array} \right.\) `(k ∈ ZZ)`