Question

Sabendo que sen^2 x/cos x = 0,45, com 0 < x < pi/2, determine o valor de sen x: 

Answer 1

$\dfrac{\mathrm{sen^2\,}x}{\cos x}=0,45\\\\\\ \dfrac{\mathrm{sen^2\,}x}{\cos x}=\dfrac{45}{100}\\\\\\ \dfrac{\mathrm{sen^2\,}x}{\cos x}=\dfrac{\diagup\!\!\!\! 5\cdot 9}{\diagup\!\!\!\! 5\cdot 20}\\\\\\ \dfrac{\mathrm{sen^2\,}x}{\cos x}=\dfrac{9}{20}\\\\\\ 20\,\mathrm{sen^2\,}x=9\cos x~~~~~~(\text{mas }\mathrm{sen^2\,}x=1-\cos^2 x)\\\\ 20\,(1-\cos^2 x)=9\cos x$

$20-20\cos^2 x=9\cos x\\\\ 0=20\cos^2 x+9\cos x-20\\\\ 20\cos^2 x+9\cos x-20=0$


Fazendo a seguinte mudança de variável:

$\cos x=t~~~~~~(-1\le t\le 1)$

a equação fica

$20t^2+9t-20=0~~~\Rightarrow~~\left\{\! \begin{array}{l} a=20\\b=9\\c=-20 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=9^2-4\cdot 20\cdot (-20)\\\\ \Delta=81+1\,600\\\\ \Delta=1\,681=41^2$


$t=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ t=\dfrac{-9\pm \sqrt{41^2}}{2\cdot 20}\\\\\\ t=\dfrac{-9\pm 41}{40}\\\\\\ \begin{array}{rcl} t=\dfrac{-9+41}{40}&~\text{ ou }~&t=\dfrac{-9-41}{40}\\\\\\ t=\dfrac{32}{40}&~\text{ ou }~&t=\dfrac{-50}{40}\\\\\\ t=\dfrac{4}{5}&~\text{ ou }~&t=-\,\dfrac{5}{4}<-1~~~~(\text{n\~ao serve}) \end{array}$


Logo, temos

$t=\dfrac{4}{5}$


Voltando à variável $x\,,$ temos

$\cos x=\dfrac{4}{5}\\\\\\ 5\cos x=4$


Elevando os dois lados ao quadrado, temos

$(5\cos x)^2=4^2\\\\ 25\cos^2 x=16~~~~~~(\text{mas }\cos^2 x=1-\mathrm{sen^2\,}x)\\\\ 25\,(1-\mathrm{sen^2\,}x)=16\\\\ 25-25\,\mathrm{sen^2\,}x=16\\\\25-16=25\,\mathrm{sen^2\,}x\\\\ 9=25\,\mathrm{sen^2\,}x\\\\ \mathrm{sen^2\,}x=\dfrac{9}{25}\\\\\\ \mathrm{sen\,}x=\pm \sqrt{\dfrac{9}{25}}\\\\\\ \mathrm{sen\,}x=\pm \dfrac{3}{5}$


Como $x$ é um arco do 1º quadrante, o seno de $x$ é positivo. Portanto,

$\mathrm{sen\,}x=\dfrac{3}{5}\\\\\\ \boxed{\begin{array}{c}\mathrm{sen\,}x=0,6 \end{array}}$