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[SPAM] [obm-l] Trigonometria

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Amigos ajude-me a entender essa solu=E7=E3o.

 Determine todos x no intervalo [0,2p] da seguinte equa=E7=E3o

         81sen^10(x) + cox^10(x) =3D 81/256

    Eu vi no forum a seguinte solu=E7=E3o:

                 se   sen^2 (x) =3D ( 1 - 3z)/4 com ( -1=3D< z =3D< =
1/3). Primeira d=FAvida como ele chegou a essa comclus=E3o? cotinuando. =
Usando a rela=E7=E3o fundamental ele encontrou cos^2(x) =3D 3.(1+z)/4 =
a=ED tudo bem.  =20

                   Ele fez o seguinte :

                                     (1 - z )^5 +3(1+z)^5 =3D4 como =
arrumo essa equa=E7=E3o?

                                   z^2(2 - 4z +7z^2- 4z^3) =3D 0

                          1. z =3D0 implica x =3D+/- (p/6) +kp  , onde p =
=3Dpi e =F3bvio que nao h=E1 outra solu=E7=E3o no inetrvalo
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2900.2180" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Amigos ajude-me a entender essa=20
solu=E7=E3o.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>&nbsp;Determine todos x no intervalo =
[0,2p] da=20
seguinte equa=E7=E3o</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial =
size=3D2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
81sen^10(x) + cox^10(x) =3D 81/256</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>&nbsp;&nbsp;&nbsp; Eu vi no forum a =
seguinte=20
solu=E7=E3o:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial=20
size=3D2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
se&nbsp;&nbsp; sen^2 (x) =3D ( 1 - 3z)/4 com ( -1=3D&lt; z =3D&lt; 1/3). =
Primeira=20
d=FAvida como ele chegou a essa comclus=E3o? cotinuando. Usando a =
rela=E7=E3o=20
fundamental ele encontrou cos^2(x) =3D 3.(1+z)/4 a=ED tudo=20
bem.&nbsp;&nbsp;&nbsp;</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial=20
size=3D2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
Ele fez o seguinte :</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial=20
size=3D2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=
&nbsp;=20
(1 - z )^5 +3(1+z)^5 =3D4 como arrumo essa equa=E7=E3o?</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial=20
size=3D2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
z^2(2 - 4z +7z^2- 4z^3) =3D 0</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial=20
size=3D2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;=20
1. z =3D0 implica x =3D+/- (p/6) +kp&nbsp; , onde p =3Dpi e =F3bvio que =
nao h=E1 outra=20
solu=E7=E3o no inetrvalo</FONT></DIV></BODY></HTML>

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