Function parameters and cartesian curves Given
$$x = \cos t + \cos 2t \,,\;  y = \sin t + \sin 2t ,$$
find the tangent line for the parameter at point $(-1, 1),$ and draw a  graph of the curve.
To find the point you could simply do $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ and find $\dfrac{dy}{dx}$ by taking the quotient, so you get $$ \dfrac{dy}{dx} =\dfrac{\cos t+2 \cos 2t}{-\sin t-2 \sin 2t}$$ 
How exactly would you define parameter $t$ in this case as a value to find the equation of the tangent? And when trying to draw the graph, you would need the Cartesian curve, which can be fetched by eliminating the parameter by solving for $t$. Is there a trigonometric identity that would make it possible to solve for $t$ using $X$ and $Y$ equations?

Another way by finding implicit equation
\begin{align*}
  x^{2} &= \cos^{2} t+2\cos t \cos 2t+ \cos^{2} 2t \\
  y^{2} &= \sin^{2} t+2\sin t \sin 2t+ \sin^{2} 2t \\
  x^{2}+y^{2} &= 2+2(\cos t \cos 2t+\sin t \sin 2t) \\
  &= 2+2\cos t \\
  \cos t &= \frac{x^{2}+y^{2}}{2}-1 \\
  x &= \cos t+2\cos^{2} t-1 \\
  &=(\cos t+1)(2\cos t-1) \\
  &= \frac{(x^{2}+y^{2})}{2} (x^{2}+y^{2}-3) \\
  0 &= (x^{2}+y^{2})(x^{2}+y^{2}-3)-2x \\
  0 &= x^{4}+2x^{2}y^{2}+y^{4}-3x^{2}-3y^{2}-2x \\
\end{align*}
In principle, the tangent can be found by implicit differentiation and plug in the point $(-1,1)$.
The rest of work is left as an exercise, but showing the graph instead:

P.S.: The curve is actually a shifted limaçon of Pascal, 
$r=2\cos \theta+1$.
