Question

9 -2cos" (x) = 15 sen (x) no intervalo [ [ - pi/2 ; pi/2 ]

Answer 1

Pela Relação Fundamental, temos:

$sen^2x+cos^2x=1\\cos^2=1-sen^2x$

Portanto,

$9-2cos^2x=15.senx\\9-2.(1-sen^2x)=15senx\\9-2.2sen^2x=15senx\\2sen^2x-15senx+7=0\\D=(-15)^2-4.(2).(7)\\D=169\\x= \frac{-(-15)+/- \sqrt{169}}{2.(2)} =\ \textgreater \ \left \{ {{x'= \frac{15+13}{4}=\ \textgreater \ x'=7(Falso) } \atop {x"= \frac{15-13}{4}=\ \textgreater \ x"= \frac{1}{2}(Verdadeiro)} \right.\\x"= \frac{1}{2}=\ \textgreater \ senx= \frac{1}{2}=\ \textgreater \ \frac{ \pi }{6}$