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Re: [obm-l] Intervalo

To: obm-l@xxxxxxxxxxxxxx
Subject: Re: [obm-l] Intervalo
From: Luiz Miletto <miletto@xxxxxxxxx>
Date: Tue, 18 Oct 2005 14:39:11 -0300
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log2 (senx) - log2 (cosx + senx) = 0

log2 (senx) = log2 (cosx + senx)
senx = cosx + senx

a) cosx =0 ===> x=pi/2 + K*pi

no intervalo [ - 2pi , 2pi]. temos: -3pi/2, -pi/2, +pi/2, +3pi/2.

b) a menor das raizes no intervalo [ - 2pi , 2pi= -3pi/2.

On 10/18/05, Anna Luisa <annaluisacg@terra.com.br> wrote:

Por favor me alguem me dah uma maozinha!

- Dada a equação log2 (senx) - log2 (cosx + senx) = 0

Determine:

a) cos x.

b) a menor raiz no intervalo [ - 2pi , 2pi].

Obrigada,

Anninha.

References:
- [obm-l] Intervalo
  - From: Anna Luisa

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