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.
 A: 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.
A: Solve t first:
$$
x=Cos(t)+1-2Sin^2(t)=-1, y=Sin(t)+2Sin(t)Cos(t)=1
$$
Which is:
$$
a+1-2b^2=-1, b+2ab=1, a^2+b^2=1
$$
So:
$$
2a^2+a=0
$$
Then I find a = 0, b=1. for a = -1/2, b cannot be solved. That is $$Cos(t)=0, Cos(2t)=-1, Sin(t)=1, Sin(2t)=0.$$
So:
$$
\frac{Cos(t)+2Cos(2t)}{-Sin(t)-2Sin(2t)}=2.
$$
A: We need $$\cos t+\cos2t=x=-1\text{ and }\sin t+\sin2t=1$$
As $\cos2t=2\cos^2t-1,$
$$2\cos^2t-1+\cos t=-1\iff\cos t(2\cos t+1)=0$$
If $\cos t=0,\sin2t=2\cos t\sin t=0\implies\sin t=1\implies t=2n\pi+\dfrac\pi2$
Now the value of the slope can be easily determined
If $2\cos t+1=0,1=\sin t+\sin2t=\sin t(1+2\cos t)=\sin t\cdot0$ which is impossible
A: I am not sure, but I suspect your title typo error.
Find the equation of tangent line to the curve $x=\cos(t)+\cos(2t), y=\sin(t)+\sin(2t) $ at the point $ (−1,0)$.
The question is about double slope of Cardoid class curve for which parametrization is given including a double point at $ (-1,0)$. ( can be also seen in Frieder's post). 
I suggest find out the two $t$s at double point  $ (x,y)= (-1,0)$ and their corresponding  two slopes as per $t$ relation you derived already. 
A: I am not sure, but I suspect your title typo error.
Find the equation of tangent line to the curve $x=\cos(t)+\cos(2t), y=\sin(t)+\sin(2t) $ at the point $ (−1,0)$.
The question is about double slope of Cardoid class curve for which parametrization is given including a double point at $ (-1,0)$  corresponding to $ t=\pi$. ( Double self -intersection point can be also seen in Frieder's post). 
I suggest you find out the two $t$s at double point  $ (x,y)= (-1,0)$ and their corresponding  two slopes as per $t$ relation you derived already. 
