da Gi8 » 29/06/2010, 14:02
In generale, $cos(alpha)=cos(beta) hArr alpha=beta + 2kpi$ $ vv $ $alpha =-beta+2kpi$, $k in ZZ$
Quindi avresti $3x+2=2x+1 + 2kpi$ $ vv $ $ 3x+2=-2x-1 +2kpi$, $k in ZZ$
Risolvendo queste, hai tutte le soluzioni del problema
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