Question

Desafio da hora
Determine m para que x=π/4, que seja raiz da equação tg^2x-m.cos^2.x+sen^2.x=0

Answer 1

Resposta:

Se diz que é raiz, basta substituir:

\begin{gathered}tg^2x-mcos^2x+sen^2x = 0\\\\ tg^2(\frac{\pi}{3})-mcos^2(\frac{\pi}{3})+sen^2(\frac{\pi}{3}) = 0\\\\ (\frac{\sqrt{3}}{3})^2-m(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2 = 0\\\\ \frac{3}{9} - m\frac{1}{4}+\frac{3}{4} = 0\\\\ \frac{1}{3}-m\frac{1}{4}+\frac{3}{4} = 0\\\\ 4-3m+9 = 0\\\\ -3m+13=0\\\\ 3m = 13\\\\ \boxed{m=\frac{13}{3}} \end{gathered}tg2x−mcos2x+sen2x=0tg2(3π)−mcos2(3π)+sen2(3π)=0(33)2−m(21)2+(23)2=093−m41+43=031−m41+43=04−3m+9=0−3m+13=03m=13m=313