If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. 
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$" 
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$  to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question 
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by  by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?

If $\sin\alpha + \cos\alpha = 0.2$, then squaring both sides and simplifying will produce $2 \sin \alpha \cos \alpha = -0.96$.
It follows that $\sin^2 \alpha - 2 \sin(\alpha) \cos(\alpha) + \cos^2 \alpha = 1.96$, from which we conclude $\sin\alpha - \cos\alpha = \pm 1.4$
From
$\left\{
    \begin{array}{l}
      \sin\alpha + \cos\alpha = 0.2\\
      \sin\alpha - \cos\alpha = 1.4
    \end{array}
\right \}$
we find $(\sin \alpha, \cos \alpha) = (0.8, -0.6)$
From
$\left\{
    \begin{array}{l}
      \sin\alpha + \cos\alpha = 0.2\\
      \sin\alpha - \cos\alpha = -1.4
    \end{array}
\right \}$
we find $(\sin \alpha, \cos \alpha) = (-0.6, 0.8)$
Note that neither of these solutions is extraneous.
