Qual é o conjunto verdade de: Sen x = sen 11pi/6
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A solução de uma equação na forma
![\mathrm{sen\,}x=\mathrm{sen\,}a \mathrm{sen\,}x=\mathrm{sen\,}a](https://tex.z-dn.net/?f=%5Cmathrm%7Bsen%5C%2C%7Dx%3D%5Cmathrm%7Bsen%5C%2C%7Da)
é
![\begin{array}{rcl} x=a+k\cdot 2\pi&\text{ ou }&x=\left(\pi-a \right )+k\cdot 2\pi \end{array} \begin{array}{rcl} x=a+k\cdot 2\pi&\text{ ou }&x=\left(\pi-a \right )+k\cdot 2\pi \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D+x%3Da%2Bk%5Ccdot+2%5Cpi%26amp%3B%5Ctext%7B+ou+%7D%26amp%3Bx%3D%5Cleft%28%5Cpi-a+%5Cright+%29%2Bk%5Ccdot+2%5Cpi+%5Cend%7Barray%7D)
onde
pertence ao conjunto dos números inteiros, isto é,
![k \in \mathbb{Z} k \in \mathbb{Z}](https://tex.z-dn.net/?f=k+%5Cin+%5Cmathbb%7BZ%7D)
Logo, para a equação proposta, temos
![\mathrm{sen\,}x=\mathrm{sen\,}\dfrac{11\pi}{6}\\ \\ \begin{array}{rcl} x=\dfrac{11\pi}{6}+k\cdot 2\pi&\text{ ou }&x=\left(\pi-\dfrac{11\pi}{6} \right )+k\cdot 2\pi \end{array}\\ \\ \\ \begin{array}{rcl} x=\dfrac{11\pi}{6}+k\cdot 2\pi&\text{ ou }&x=\dfrac{6\pi-11\pi}{6}+k\cdot 2\pi \end{array}\\ \\ \\ \begin{array}{rcl} x=\dfrac{11\pi}{6}+k\cdot 2\pi&\text{ ou }&x=-\dfrac{5\pi}{6}+k\cdot 2\pi \end{array} \mathrm{sen\,}x=\mathrm{sen\,}\dfrac{11\pi}{6}\\ \\ \begin{array}{rcl} x=\dfrac{11\pi}{6}+k\cdot 2\pi&\text{ ou }&x=\left(\pi-\dfrac{11\pi}{6} \right )+k\cdot 2\pi \end{array}\\ \\ \\ \begin{array}{rcl} x=\dfrac{11\pi}{6}+k\cdot 2\pi&\text{ ou }&x=\dfrac{6\pi-11\pi}{6}+k\cdot 2\pi \end{array}\\ \\ \\ \begin{array}{rcl} x=\dfrac{11\pi}{6}+k\cdot 2\pi&\text{ ou }&x=-\dfrac{5\pi}{6}+k\cdot 2\pi \end{array}](https://tex.z-dn.net/?f=%5Cmathrm%7Bsen%5C%2C%7Dx%3D%5Cmathrm%7Bsen%5C%2C%7D%5Cdfrac%7B11%5Cpi%7D%7B6%7D%5C%5C+%5C%5C+%5Cbegin%7Barray%7D%7Brcl%7D+x%3D%5Cdfrac%7B11%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%26amp%3B%5Ctext%7B+ou+%7D%26amp%3Bx%3D%5Cleft%28%5Cpi-%5Cdfrac%7B11%5Cpi%7D%7B6%7D+%5Cright+%29%2Bk%5Ccdot+2%5Cpi+%5Cend%7Barray%7D%5C%5C+%5C%5C+%5C%5C+%5Cbegin%7Barray%7D%7Brcl%7D+x%3D%5Cdfrac%7B11%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%26amp%3B%5Ctext%7B+ou+%7D%26amp%3Bx%3D%5Cdfrac%7B6%5Cpi-11%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi+%5Cend%7Barray%7D%5C%5C+%5C%5C+%5C%5C+%5Cbegin%7Barray%7D%7Brcl%7D+x%3D%5Cdfrac%7B11%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%26amp%3B%5Ctext%7B+ou+%7D%26amp%3Bx%3D-%5Cdfrac%7B5%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi+%5Cend%7Barray%7D)
onde![k \in \mathbb{Z} k \in \mathbb{Z}](https://tex.z-dn.net/?f=k+%5Cin+%5Cmathbb%7BZ%7D)
O conjunto verdade para esta equação é
![S=\left\{x \in \mathbb{R}\left|\,x=\dfrac{11\pi}{6}+k\cdot 2\pi\;\text{ ou }\;x=-\dfrac{5\pi}{6}+k\cdot 2\pi,\;k \in \mathbb{Z}\right. \right \} S=\left\{x \in \mathbb{R}\left|\,x=\dfrac{11\pi}{6}+k\cdot 2\pi\;\text{ ou }\;x=-\dfrac{5\pi}{6}+k\cdot 2\pi,\;k \in \mathbb{Z}\right. \right \}](https://tex.z-dn.net/?f=S%3D%5Cleft%5C%7Bx+%5Cin+%5Cmathbb%7BR%7D%5Cleft%7C%5C%2Cx%3D%5Cdfrac%7B11%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%5C%3B%5Ctext%7B+ou+%7D%5C%3Bx%3D-%5Cdfrac%7B5%5Cpi%7D%7B6%7D%2Bk%5Ccdot+2%5Cpi%2C%5C%3Bk+%5Cin+%5Cmathbb%7BZ%7D%5Cright.+%5Cright+%5C%7D)
é
onde
Logo, para a equação proposta, temos
onde
O conjunto verdade para esta equação é
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