resolva à equação: raiz quadrada 3 sin-Cos x=1
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Resolver a equação trigonométrica:
![\mathsf{\sqrt{3}\,sen\,x-cos\,x=1} \mathsf{\sqrt{3}\,sen\,x-cos\,x=1}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Csqrt%7B3%7D%5C%2Csen%5C%2Cx-cos%5C%2Cx%3D1%7D)
Multiplicando os dois lados da equação por![\mathsf{\dfrac{1}{2}:} \mathsf{\dfrac{1}{2}:}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%3A%7D)
![\mathsf{\dfrac{1}{2}\cdot
\left(\sqrt{3}\,sen\,x-cos\,x\right )=\dfrac{1}{2}}\\\\\\
\mathsf{\dfrac{\sqrt{3}}{2}\,sen\,x-\dfrac{1}{2}\,cos\,x=\dfrac{1}{2}\qquad(i)} \mathsf{\dfrac{1}{2}\cdot
\left(\sqrt{3}\,sen\,x-cos\,x\right )=\dfrac{1}{2}}\\\\\\
\mathsf{\dfrac{\sqrt{3}}{2}\,sen\,x-\dfrac{1}{2}\,cos\,x=\dfrac{1}{2}\qquad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%0A+%5Cleft%28%5Csqrt%7B3%7D%5C%2Csen%5C%2Cx-cos%5C%2Cx%5Cright+%29%3D%5Cdfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5C+%0A%5Cmathsf%7B%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5C%2Csen%5C%2Cx-%5Cdfrac%7B1%7D%7B2%7D%5C%2Ccos%5C%2Cx%3D%5Cdfrac%7B1%7D%7B2%7D%5Cqquad%28i%29%7D)
Mas,
![\begin{cases}
\mathsf{\dfrac{\sqrt{3}}{2}=cos\,\dfrac{\pi}{6}}\\\\
\mathsf{\dfrac{1}{2}=sen\,\dfrac{\pi}{6}}
\end{cases}\qquad\quad\left(\textsf{lembre que
}\mathsf{\dfrac{\pi}{6}=30^\circ}\right) \begin{cases}
\mathsf{\dfrac{\sqrt{3}}{2}=cos\,\dfrac{\pi}{6}}\\\\
\mathsf{\dfrac{1}{2}=sen\,\dfrac{\pi}{6}}
\end{cases}\qquad\quad\left(\textsf{lembre que
}\mathsf{\dfrac{\pi}{6}=30^\circ}\right)](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%0A+%5Cmathsf%7B%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3Dcos%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D%7D%5C%5C%5C%5C+%0A%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%3Dsen%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D%7D+%0A%5Cend%7Bcases%7D%5Cqquad%5Cquad%5Cleft%28%5Ctextsf%7Blembre+que+%0A%7D%5Cmathsf%7B%5Cdfrac%7B%5Cpi%7D%7B6%7D%3D30%5E%5Ccirc%7D%5Cright%29)
Substituindo em
ficamos com
![\mathsf{cos\,\dfrac{\pi}{6}\,sen\,x-sen\,\dfrac{\pi}{6}\,cos
x=\dfrac{1}{2}}\\\\\\\
\mathsf{sen\,x\,cos\,\dfrac{\pi}{6}-cos\,x\,sen\,\dfrac{\pi}{6}=\dfrac{1}{2}\qquad(ii)} \mathsf{cos\,\dfrac{\pi}{6}\,sen\,x-sen\,\dfrac{\pi}{6}\,cos
x=\dfrac{1}{2}}\\\\\\\
\mathsf{sen\,x\,cos\,\dfrac{\pi}{6}-cos\,x\,sen\,\dfrac{\pi}{6}=\dfrac{1}{2}\qquad(ii)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bcos%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D%5C%2Csen%5C%2Cx-sen%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D%5C%2Ccos%0A+x%3D%5Cdfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5C%5C+%0A%5Cmathsf%7Bsen%5C%2Cx%5C%2Ccos%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D-cos%5C%2Cx%5C%2Csen%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Cqquad%28ii%29%7D)
Lembremos da fómula do seno da diferença entre dois arcos:
![\mathsf{sen(\alpha-\beta)=sen\,\alpha\,cos\,\beta-cos\,\alpha\,sen\,\beta} \mathsf{sen(\alpha-\beta)=sen\,\alpha\,cos\,\beta-cos\,\alpha\,sen\,\beta}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsen%28%5Calpha-%5Cbeta%29%3Dsen%5C%2C%5Calpha%5C%2Ccos%5C%2C%5Cbeta-cos%5C%2C%5Calpha%5C%2Csen%5C%2C%5Cbeta%7D)
Podemos notar que o lado esquerdo de
é o seno da diferença, onde
![\mathsf{\alpha=x~~e~~\beta=\dfrac{\pi}{6}.} \mathsf{\alpha=x~~e~~\beta=\dfrac{\pi}{6}.}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Calpha%3Dx%7E%7Ee%7E%7E%5Cbeta%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D.%7D)
Então, a equação
fica
![\mathsf{sen\!\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}}\\\\\\
\mathsf{sen\!\left(x-\dfrac{\pi}{6}\right)=sen\,\dfrac{\pi}{6}} \mathsf{sen\!\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}}\\\\\\
\mathsf{sen\!\left(x-\dfrac{\pi}{6}\right)=sen\,\dfrac{\pi}{6}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsen%5C%21%5Cleft%28x-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Cright%29%3D%5Cdfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5C%0A+%5Cmathsf%7Bsen%5C%21%5Cleft%28x-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Cright%29%3Dsen%5C%2C%5Cdfrac%7B%5Cpi%7D%7B6%7D%7D)
Agora temos uma igualdade entre senos. Portanto, devemos ter
![\begin{array}{rcl}
\mathsf{x-\dfrac{\pi}{6}}=\dfrac{\pi}{6}+k\cdot 2\pi&~\textsf{ ou
}~&\mathsf{x-\dfrac{\pi}{6}=\left(\pi-\dfrac{\pi}{6}\right)+k\cdot
2\pi}\\\\ \mathsf{x=\dfrac{\pi}{6}+\dfrac{\pi}{6}+k\cdot
2\pi}&~\textsf{ ou
}~&\mathsf{x=\left(\pi-\dfrac{\pi}{6}\right)+\dfrac{\pi}{6}+k\cdot
2\pi}\\\\ \mathsf{x=\dfrac{2\pi}{6}+k\cdot 2\pi}&~\textsf{ ou
}~&\mathsf{x=\pi-\dfrac{\pi}{6}+\dfrac{\pi}{6}+k\cdot 2\pi}\\\\
\mathsf{x=\dfrac{\pi}{3}+k\cdot 2\pi}&~\textsf{ ou
}~&\mathsf{x=\pi+k\cdot 2\pi} \end{array} \begin{array}{rcl}
\mathsf{x-\dfrac{\pi}{6}}=\dfrac{\pi}{6}+k\cdot 2\pi&~\textsf{ ou
}~&\mathsf{x-\dfrac{\pi}{6}=\left(\pi-\dfrac{\pi}{6}\right)+k\cdot
2\pi}\\\\ \mathsf{x=\dfrac{\pi}{6}+\dfrac{\pi}{6}+k\cdot
2\pi}&~\textsf{ ou
}~&\mathsf{x=\left(\pi-\dfrac{\pi}{6}\right)+\dfrac{\pi}{6}+k\cdot
2\pi}\\\\ \mathsf{x=\dfrac{2\pi}{6}+k\cdot 2\pi}&~\textsf{ ou
}~&\mathsf{x=\pi-\dfrac{\pi}{6}+\dfrac{\pi}{6}+k\cdot 2\pi}\\\\
\mathsf{x=\dfrac{\pi}{3}+k\cdot 2\pi}&~\textsf{ ou
}~&\mathsf{x=\pi+k\cdot 2\pi} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%0A+%5Cmathsf%7Bx-%5Cdfrac%7B%5Cpi%7D%7B6%7D%7D%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%26amp%3B%7E%5Ctextsf%7B+ou+%0A%7D%7E%26amp%3B%5Cmathsf%7Bx-%5Cdfrac%7B%5Cpi%7D%7B6%7D%3D%5Cleft%28%5Cpi-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Cright%29%2Bk%5Ccdot+%0A2%5Cpi%7D%5C%5C%5C%5C+%5Cmathsf%7Bx%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%2B%5Cdfrac%7B%5Cpi%7D%7B6%7D%2Bk%5Ccdot+%0A2%5Cpi%7D%26amp%3B%7E%5Ctextsf%7B+ou+%0A%7D%7E%26amp%3B%5Cmathsf%7Bx%3D%5Cleft%28%5Cpi-%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Cright%29%2B%5Cdfrac%7B%5Cpi%7D%7B6%7D%2Bk%5Ccdot+%0A2%5Cpi%7D%5C%5C%5C%5C+%5Cmathsf%7Bx%3D%5Cdfrac%7B2%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%7D%26amp%3B%7E%5Ctextsf%7B+ou+%0A%7D%7E%26amp%3B%5Cmathsf%7Bx%3D%5Cpi-%5Cdfrac%7B%5Cpi%7D%7B6%7D%2B%5Cdfrac%7B%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%7D%5C%5C%5C%5C+%0A%5Cmathsf%7Bx%3D%5Cdfrac%7B%5Cpi%7D%7B3%7D%2Bk%5Ccdot+2%5Cpi%7D%26amp%3B%7E%5Ctextsf%7B+ou+%0A%7D%7E%26amp%3B%5Cmathsf%7Bx%3D%5Cpi%2Bk%5Ccdot+2%5Cpi%7D+%5Cend%7Barray%7D)
onde
é um inteiro.
Conjunto solução:![\mathsf{S=\left\{x\in\mathbb{R}:~~x=\dfrac{\pi}{3}+k\cdot
2\pi~~ou~~x=\pi+k\cdot 2\pi,~~k\in\mathbb{Z}\right\}} \mathsf{S=\left\{x\in\mathbb{R}:~~x=\dfrac{\pi}{3}+k\cdot
2\pi~~ou~~x=\pi+k\cdot 2\pi,~~k\in\mathbb{Z}\right\}}](https://tex.z-dn.net/?f=%5Cmathsf%7BS%3D%5Cleft%5C%7Bx%5Cin%5Cmathbb%7BR%7D%3A%7E%7Ex%3D%5Cdfrac%7B%5Cpi%7D%7B3%7D%2Bk%5Ccdot+%0A2%5Cpi%7E%7Eou%7E%7Ex%3D%5Cpi%2Bk%5Ccdot+2%5Cpi%2C%7E%7Ek%5Cin%5Cmathbb%7BZ%7D%5Cright%5C%7D%7D)
Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7489224
Dúvidas? Comente.
Bons estudos! :-)
Tags: equação trigonométrica seno sen cosseno cos identidade diferença soma arcos solução
Multiplicando os dois lados da equação por
Mas,
Substituindo em
Lembremos da fómula do seno da diferença entre dois arcos:
Podemos notar que o lado esquerdo de
Então, a equação
Agora temos uma igualdade entre senos. Portanto, devemos ter
onde
Conjunto solução:
Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7489224
Dúvidas? Comente.
Bons estudos! :-)
Tags: equação trigonométrica seno sen cosseno cos identidade diferença soma arcos solução
wallace106:
vlw mais conseguir fazer do mesmo jeito, obg pela ajuda
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