<s>[INST] 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
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
 [/INST] $$\cos \alpha + \sin \alpha = \sqrt{2} (\frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}) = \sqrt{2}(\sin \frac{\pi}{4} \cos\alpha +  \cos \frac{\pi}{4} \sin \alpha) 
= \sqrt{2}\sin(\frac{\pi}{4} + \alpha) = .2 $$ 
Taking the inverse sin of each side yields
$$ \alpha = 2.2143 + 2 \pi n_1 \mid n_1 \in \mathbb{Z} \,\,\,\text{or}\,\,\, \alpha = 2\pi n_2 - .643501 \mid n_2 \in \mathbb{Z}$$
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