1 Exercit˘ii rezolvate - Deliu - 2cos 2 2x cos2x cos6x

2 (cos8x+cos6x) = 1 2 (cos8x+cos2x) ⇔ cos6x−cos2x=0 ⇔ −2sin4xsin2x=0 sin4x=0 ⇒ x=kˇ 4;k∈Z sin2x=0 ⇒ x=kˇ 2;k∈Z, mult˘ime de solut˘ii care este inclus a ^ n prima mult˘ime. 2 Tem a 1.cosx+ √ 3sinx=m; discut˘ie dup a m∈R 2.2cos2 x−sin2x+sinx+cosx=1 3.cos2 x+3sin2 x+2 √ 3sinxcosx=1 4.cos2 x+cos2 2x+cos2 3x+cos2 4x=2 5 ... ................
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