Question

determinar o valor do cos(2x) sabendo que x satisfaz senx + cos =2/3 (dois sobre três)

Answer 1

$senx + cosx= \dfrac{2}{3}$

$(senx + cosx)^{2} = (\dfrac{2}{3})^{2}$

$sen^{2} x+2senxcosx+ cos^{2} x=\dfrac{4}{9}$

$sen^{2} x+ cos^{2} x+2senxcosx=\dfrac{4}{9}$

$1+sen2x=\dfrac{4}{9}$

$sen2x=\dfrac{4}{9}-1=\dfrac{4-9}{9}=-\dfrac{5}{9}$

$sen^{2} (2x)+cos^{2} (2x)=1$

$(-\dfrac{5}{9})^{2}+cos^{2} (2x)=1$

$\dfrac{25}{81}+cos^{2} (2x)=1$

$cos^{2} (2x)=1-\dfrac{25}{81}$

$cos^{2} (2x)=\dfrac{81-25}{81}$

$cos^{2} (2x)=\dfrac{56}{81}$

$cos(2x)= \sqrt{ \dfrac{56}{81}}= \dfrac{ 2\sqrt{14} }{9}$
ou
$cos(2x)=- \sqrt{ \dfrac{56}{81}}= -\dfrac{ 2\sqrt{14} }{9}$