Solving  $\cos^2 \theta + \cos \theta = 2$ Solve the following for $\theta$:  

$\cos^2 \theta + \cos \theta = 2$  [Hint: There is only one solution.]

I started this out by changing $\cos^2\theta$ to $\dfrac{1+\cos(2\theta)}{2}+\cos\theta=2$
$1+\cos(2\theta)$ turns into $1+\cos^2\theta-\sin^2\theta$ which all becomes; $\dfrac{1+\cos^2\theta-\sin^2\theta}{2}+\cos\theta=2$
  Not to sure what to do after this. I was going to try a power reducing rule for $\sin^2\theta$ but that would make $\dfrac{1+\cos^2\theta- \left(\frac{1-\cos(2\theta)}2 \right)}2+\cos\theta=2$. Please do help. 

Replacing $\cos^2\theta$ with and expression involving $\cos2\theta$ is not necessarily a good idea; then you have to deal with cosines of two different angles.
A better approach is to realize that what we have is a quadratic equation: let us define $y$ to be $y=\cos\theta$. Then we can rewrite the equation as
$$y^2 + y = 2$$
or $y^2 + y - 2 = 0$.
We know how to solve quadratic equations: the solutions are
$$\begin{align*}
y_1 &= \frac{-1+\sqrt{1+8}}{2} = \frac{-1+3}{2} = 1\\
y_2 &= \frac{-1-\sqrt{1+8}}{2} = \frac{-1-3}{2} = -2.
\end{align*}$$
However, now we remember that $y$ is actually $\cos\theta$, so now we want to find the solutions to $\cos\theta = 1$ and of $\cos\theta=-2$.
Since $-1\leq\cos\theta\leq 1$, the latter equation has no solutions. 
So the answer is that the solutions are exactly the $\theta$ for which $\cos(\theta)=1$.
(Which we could have figured out cleverly by  making the observation made by David Mitra in comments, but I wanted to give you an idea of how to approach this kind of equation if the answer is not so obvious.)
