2 ESERCIZI Equazioni e disequazioni goniometriche esercizio svolto tan2 2x + tan2x 6 = 0 3 _ 1 1 + 24 1 5 tan2x = ____________ = ______ = 2 2 2 Abbiamo: tan2x = 3 tan2x = 2 1 x = __ arctan3 + k __ 2 2 1 x = __ arctan2 + k __ 2 2 2x = arctan( 3) + k 2x = arctan2 + k 65 sen2y 2seny + 1 = 0 66 4cos2y 4 3cosy + 3 = 0 67 cos24x 3cos4x + 2 = 0 68 tan2x (1 + 3)tanx + 3 = 0 69 tan22x + tan2x + 2 = 0 _ _ [ 2 + 2k ] 75 sen43y 2sen23y + 1 = 0 _ _ [ 6 + 2k ] 76 3sen2z cos2z = 0 77 2sen2x cosx 1 = 0 _ _ [ + 2k ; 3 + 2k ] [ 4 + k ; 3 + k ] 78 2cos2y + seny 1 = 0 [ 2 + 2k ; 2 3 + 2k ] [ ] 79 7 + 113 2sen2 7cos + 6 = 0 [ arccos___________ + 2k ] __ __ _ _ [0; (k 2)] __ _ _ _ _ __ 2 _x_ x 70 2cos + 2 2cos __ + 1 = 0 2 2 _ _ [2 2 + k4 ] 71 2sen2 3sen + 1 = 0 72 cosy sen2y = 1 73 1 3tan22y + 2tan2y = 0 [_ _ + k _ _; _1_ arctan( _1_) + k _ _] 74 cos22 4cos2 + 1 = 0 [ _1_ arccos(2 3 ) + k ] 2 _ _ _ _ _ _ [ 2 + 2k ; 2 3 + 2k ] [0 + 2k ] 8 2 2 3 2 __ _ _ _ _ [6 + k 3] _ _ [ 6 + k ] _3_ _ _ _ _ _ 80 2 __ __ tan 2x 2tan2x + 2 1 = 0 1 tanx ____ = 0 senx 1 82 tan2x + _____ =0 cos2x 1 83 tan2x + 1 = _____ cos2x 81 84 4 _ _ _ _ ___ __ [ 8 + k 2 ; 16 + k 2 ] [ 0,90 + 2k ] [ ] __ [per ogni x 2 + k ] sen25 + cos25 1 = 0 [R] esercizio svolto x (sen3x 1) (cos __ + 1) = 0 2 Per la legge di annullamento del prodotto abbiamo: sen3x 1 = 0 x cos __ + 1 = 0 2 2 3x = __ + 2k x = __ + __ k 2 6 3 x x cos __ = 1 __ = + 2k x = 2 + 4k 2 2 sen3x = 1 _3_ _ _ _ _ 85 (sen2x 1)(sen4x + 1) = 0 [ 4 + k ; 8 (k 2)] 86 (2senx + 1)(2 cosx + 3) = 0 [ 2 3 + 2k ] 87 (2sen + 2)( cos 2) = 0 88 senx + senxcosx = 0 __ __ 89 sen __ + sen __ cos __ = 0 4 4 4 _3_ _ _ 90 3senxcosx 2cos2x = 0 [ 2 4 + 2k ] _3_ _ _ 91 tan23 tan3 = 0 [0 + k ] 92 2sen2x + senx = 0 [0 + k4 ] _ _ _1_ [ 2 + k ; arctan 3 + k ] _ _ ___ _ _ [0 + k 3 ; 12 + k 3 ] 3 __ __ [0 + k ; 2 3 + 2k ] 99