Find the equation of tangent line to the curve $x=\cos(t) + \cos(2t)$, $y= \sin(t) + \sin(2t)$ at the point $(-1,1)$. I get the equation for the slope as $\frac{\cos(t) + 2\cos(2t)}{-\sin(t) -2\sin(2t)}$ but I'm unsure how to solve for the value of $t$. I know I need to sub in $-1$ and $1$ for $x$ and $y$ but the equations are hard to solve.

Given is curve:
$$\begin{gathered}
  x(t) = \cos (t) + \cos (2t) \hfill \\
  y(t) = \sin (t) + \sin (2t) \hfill \\ 
\end{gathered}$$
$$\begin{gathered}
  \varphi (t) = \left( {\begin{array}{*{20}{c}}
  {x(t)} \\ 
  {y(t)} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\cos (t) + \cos (2t)} \\ 
  {\sin (t) + \sin (2t)} 
\end{array}} \right) \hfill \\
  \frac{{d\varphi }}{{dt}}(t) = \left( {\begin{array}{*{20}{c}}
  { - \sin (t) - 2\sin (2t)} \\ 
  {\cos (t) + 2\cos (2t)} 
\end{array}} \right) \hfill \\ 
\end{gathered}$$
From graphs for sin and cos we find:

$$\begin{gathered}
  \cos (\frac{\pi }{2}) = 0 \wedge \cos (\pi ) =  - 1 \hfill \\
  \sin (\frac{\pi }{2}) = 1 \wedge \sin (\pi ) = 0 \hfill \\ 
\end{gathered} $$
$$\begin{gathered}
  \varphi (\frac{\pi }{2}) = \left( {\begin{array}{*{20}{c}}
  { - 1} \\ 
  1 
\end{array}} \right) \hfill \\
  \frac{{d\varphi }}{{dt}}(\frac{\pi }{2}) = \left( {\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 2} 
\end{array}} \right) \hfill \\ 
\end{gathered}$$
and calculate slope:
$$\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{1}{{\frac{{dx}}{{dt}}}} =  - 2 \cdot \frac{1}{{ - 1}} = 2$$

Remark: Our tangent vector is normed and therfore has lenght equal to one. So it's
$$\frac{{d\varphi }}{{dt}} = \frac{1}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}}
  { - 1} \\ 
  { - 2} 
\end{array}} \right)$$
Slope doesn't change, because of $\frac{1}{{\sqrt 5 }}$ cancels by division.
