Respuesta :
[tex]\bf 2sin(t)cos(t)-cos(t)+2sin(t)-1=0
\\\\\\\
[2sin(t)cos(t)+2sin(t)]\quad -\quad [cos(t)+1]=0\\
\left. \qquad \right.\uparrow \\
\textit{common factoring}
\\\\\\
2sin(t)[\underline{cos(t)+1}]\quad -\quad [\underline{cos(t)+1}]=0\\
\left. \qquad \qquad \right.\uparrow \qquad \qquad \qquad\qquad \uparrow \\
\left. \qquad \right.\textit{some more common factor}
[/tex]
[tex]\bf [cos(t)+1][2sin(t)-1]=0\implies \begin{cases} 2sin(t)-1=0\\\\ sin(t)=\frac{1}{2}\\\\ \measuredangle t=sin^{-1}\left( \frac{1}{2} \right)\\\\ \measuredangle t=\frac{\pi }{6}\ ,\ \frac{5\pi }{6}\\ ----------\\ cos(t)+1=0\\\\ cos(t)=-1\\\\ \measuredangle t=cos^{-1}(-1)\\\\ \measuredangle t=\pi \end{cases}[/tex]
[tex]\bf [cos(t)+1][2sin(t)-1]=0\implies \begin{cases} 2sin(t)-1=0\\\\ sin(t)=\frac{1}{2}\\\\ \measuredangle t=sin^{-1}\left( \frac{1}{2} \right)\\\\ \measuredangle t=\frac{\pi }{6}\ ,\ \frac{5\pi }{6}\\ ----------\\ cos(t)+1=0\\\\ cos(t)=-1\\\\ \measuredangle t=cos^{-1}(-1)\\\\ \measuredangle t=\pi \end{cases}[/tex]
