Question

Alguém me ajuda por favor
Dada a expressão: sen(x)=(4k−3)/9 e cos(x)=(3k−3)/3, ache o valor de k

Answer 1

$sen^2(x)+cos^2(x)=1\\\\ ( \frac{4k-3}{9} )^2 + (\frac{3k-3}{3})^2 = 1\\\\ \frac{(4k-3)^2}{9^2} + \frac{(3k-3)^2}{3^2} =1\\\\ \frac{(16k^2-24k+9)}{9^2}+ \frac{(9k^2-18k+9)}{9} = 1\\\\ \frac{(16k^2-24k+9)}{9^2}+ \frac{(9k^2-18k+9)*9}{9*9} = 1\\\\ \frac{(16k^2-24k+9)}{9^2}+ \frac{(81k^2-162k+81)}{9^2} = 1\\\\ \frac{16k^2-24k+9+81k^2-162k+81}{9^2} =1\\\\ 97k^2-186k+90=81\\\\ 97k^2-186k+9=0$

aplica bhaskara
$97k^2-186k+9=0\\\\ k= \frac{-(-186)\pm \sqrt{(-186)^2-4*97*9} }{2*97} \\\\k= \frac{186\pm \sqrt{31104} }{2*97} \\\\k= \frac{186\pm 72\sqrt{6} }{2*97}\\\\ \boxed{\boxed{k = \frac{93\pm36 \sqrt{6} }{97} }}$