1 Exercit˘ii rezolvate - Deliu - Cos2x 4sinx 2cosx 3

1.cos2x+4sinx−1 =0 R: ^Inlocuim ^ n ecuat˘ie cos2x=1−2sin2 x˘si obt˘inem 1−2sin2 x+4sinx−1 =0 ⇔ 2sinx(2−sinx) =0 Cum sinx≤1 ⇒ 2−sinx≥0 deci singura solut˘ie acceptabil a este sinx=0 de unde obt˘inem ... 3.4sinx+2cosx−3tgx−2 =0 R: Facem substitut˘ia t=tg x 2. Avem sinx= 2t ................
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