Matemática, perguntado por gerlanmatfis, 1 ano atrás

questões de trigonometria

Anexos:

JBRY: Esta dividindo as perguntas?
gerlanmatfis: como assim?
JBRY: A ideia que o Lucas deu ai agora.
gerlanmatfis: professor eu não sei como dividir já tentei e não conseguir
JBRY: Escreva uma ou duas em cada folha e poste novamente com certeza vai ser resolvido mais rápido.
gerlanmatfis: vou tentar
gerlanmatfis: professor vai lá no meu perfil eu coloquei duas
JBRY: Posta os outros exercício separados.
gerlanmatfis: http://brainly.com.br/tarefa/4463927
gerlanmatfis: professor responde essas

Soluções para a tarefa

Respondido por JBRY
1
Boa tarde Gerlan!

Solução!

Exercicio 83

5sec(x)-3tan^{2}x=1\\\\\\
5 \dfrac{1}{cos(x)}-3 \dfrac{sen^{2}(x) }{cos^{2} (x) }=1\\\\\\
 \dfrac{5cos(x)}{cos^{2}(x) }  -3 \dfrac{sen^{2}(x) }{cos^{2} (x) }= \dfrac{cos ^{2}(x) }{cos^{2}(x)} \\\\\\\
 \dfrac{ [5cos(x)-3sen^{2}x-cos^{2}x]}{cos^{2} (x)} \\\\\\\
 \dfrac{5cos(x)-3[1-cos^{2}x]-cos^{2}x}{  cos^{2}(x) } \\\\\\\
  \dfrac{5cos(x)-3+3cos^{2}x-cos^{2}x }{2cos^{2}x }\\\\\\\
 2cos^{2}x+5cos(x)\dfrac{-3cos^{2}x }{cos^{2}x } \\\\\\\
2cos^{2}x+5cos(x)-3=0\\\\
Uma ~~equacao~~ do ~~segundo ~~grau.

2cos^{2}x+5cos(x)-3=0\\\\

Fazendo \\\\
2t^{2}+5t-3=0\\\\\\
Formula ~~de~~ Baskara.\\\\\\
t= \dfrac{-b\pm \sqrt{b^{2} -4.a.c}}{2.a} \\\\\\\
t= \dfrac{-5\pm \sqrt{5^{2} -4.2.(-3)}}{2.2} \\\\\\\
t= \dfrac{-5\pm \sqrt{25 +24}}{4} \\\\\\\
t= \dfrac{-5\pm \sqrt{49}}{4} \\\\\\\
t= \dfrac{-5\pm 7}{4} \\\\\\\
 t_{1} = \dfrac{-5+7}{4}= \dfrac{2}{4}= \dfrac{1}{2}\\\\\\\
t_{2} = \dfrac{-5-7}{4}= \dfrac{-12}{4}= -3\\\\\\\
Nao ~~existe~~ cos~~ de -3

Usando~~ a~~ relacao~~fundamental\\\\\\
sen^{2}x+cos^{2}x=1\\\\\\
sen^{2}x+ (\frac{1}{2})^{2}=1  \\\\\\\\\
sen^{2}x+ (\frac{1}{4})=1 \\\\\\
4sen^{2}x+1=4\\\\\\\
4sen^{2}x=4-1\\\\\\\
4sen^{2}x=3\\\\\\\
sen(x)= \frac{\sqrt{3} }{\sqrt{4} }\\\\\\
sen(x)= \frac{\sqrt{3} }{2}

Exercicio 84

sen^{2}(x)-5sen(x).cos(x)+cos^{2}(x)=3\\\\\
sen^{2}(x)+cos^{2}(x)-5sen(x).cos(x)=3\\\\\\
relacao~~fundamental\\\\\\ sen^{2}x+cos^{2}x=1\\\\\\\
1-5sen(x).cos(x)=3\\\\\\\\
identidade~~cos(x)tan(x)=sen(x)~~entao.\\\\\\\\
1-5sen(x).cos(x)=3\\\\\\
-5sen(x).cos(x)=3-1\\\\\\
-5sen(x).cos(x)=2\\\\\\

-5tan(x)cos(x).cos(x)=2\\\\\\
Tan(x)= \dfrac{sen(x)}{cos(x)} \\\\\\
-5 \frac{sen(x)}{cos(x)} .cos(x)=2\\\\\\
cos(x)[ \frac{-5sen(x).1}{cos(x)}]=2\\\\\\\

-5sen(x)=2\\\\\\
sen(x)= \dfrac{2}{5}\\\\\\\

Novamente~~a~~relacao~~fundamental\\\\\\
sen^{2}x+cos^{2}x=1\\\\\\\
( \frac{-2}{5}) ^{2}+cos^{2}x=1\\\\\\\
( \frac{4}{25}) ^{2}+cos^{2}x=1\\\\\\\
4+25cos^{2}x=25\\\\\\
25cos^{2}=25-4\\\\\\
 25cos^{2}=21\\\\\\
cos^{2}x= \dfrac{21}{25}\\\\\\
cos(x)= \sqrt{\dfrac{21}{25} }\\\\\\\
cos(x)= \dfrac{ \sqrt{21} }{5}\\\\\\\\\
sen(x)= \dfrac{2}{5}\\\\\\\
cos(x)= \dfrac{ \sqrt{21} }{5}\\\\\\\
Tan(x)=\dfrac{sen(x)}{cos(x)}\\\\\\\
Tan(x)=\dfrac{  \dfrac{2}{5} }{ \dfrac{ \sqrt{21}}{5}}\\\\\\\
Tan(x)=\dfrac{2}{5}\times  \dfrac{5}{\sqrt{21}} \\\\\\\

Tan(x)=\dfrac{2}{ \sqrt{21} }\\\\\\\
Tan(x)= \frac{2 \sqrt{21} }{21}

Boa tarde!

Bons estudos!
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