The image contains a question about determining the equation of a tangent line to a curve and graphing it. The vector equation is given as:

$\mathbf{r}(t) = \left( \cos t + \cos 2t \right) \mathbf{i} + \left( \sin t + \sin 2t \right) \mathbf{j}$

The task is to find the tangent line to the curve at the point (-1, -1), where $\mathbf{r}(t)$ represents a parametric curve.

To approach the problem:

1. Differentiate $\mathbf{r}(t)$ to find the tangent vector. [ \mathbf{r}'(t) = \frac{d}{dt}\left[\cos t + \cos 2t \right]\mathbf{i} + \frac{d}{dt}\left[\sin t + \sin 2t \right]\mathbf{j} ]

   [ \mathbf{r}'(t) = \left[-\sin t - 2\sin 2t \right]\mathbf{i} + \left[\cos t + 2\cos 2t \right]\mathbf{j} ]

2. Find the value of $t$ that gives the point (-1, -1). This is solved by setting the parametric equations equal to the given coordinates:

   • $\cos t + \cos 2t = -1$
   • $\sin t + \sin 2t = -1$

   Solving this system will provide the value of $t$ at which the tangent is to be calculated.

3. Substitute the value of $t$ into $\mathbf{r}'(t)$ to get the tangent vector at that point.

4. Write the parametric equation of the tangent line using the point (-1, -1) and the tangent vector found in step 3.

Would you like to go through the complete step-by-step solution?