Question

2sintcost-cost+2sint-1=0 in an interval from [0, 2pi]

Answer 1

$\bf 2sin(t)cos(t)-cos(t)+2sin(t)-1=0 \\\\\\\ [2sin(t)cos(t)+2sin(t)]\quad -\quad [cos(t)+1]=0\\ \left. \qquad \right.\uparrow \\ \textit{common factoring} \\\\\\ 2sin(t)[\underline{cos(t)+1}]\quad -\quad [\underline{cos(t)+1}]=0\\ \left. \qquad \qquad \right.\uparrow \qquad \qquad \qquad\qquad \uparrow \\ \left. \qquad \right.\textit{some more common factor}$

$\bf [cos(t)+1][2sin(t)-1]=0\implies \begin{cases} 2sin(t)-1=0\\\\ sin(t)=\frac{1}{2}\\\\ \measuredangle t=sin^{-1}\left( \frac{1}{2} \right)\\\\ \measuredangle t=\frac{\pi }{6}\ ,\ \frac{5\pi }{6}\\ ----------\\ cos(t)+1=0\\\\ cos(t)=-1\\\\ \measuredangle t=cos^{-1}(-1)\\\\ \measuredangle t=\pi \end{cases}$