Solution to trigonometric derivative Version 2
For
\begin{align}
&x(t)\text{:=}\cos (t)+\cos (2 t)+1&\\
&y(t)\text{:=}\sin (t)+\sin (2 t)&\\
\end{align}
how would I go about proving that the solutions to
\begin{align}
t\in\mathbb{R}:\frac{2 \left(x'(t)^2+y'(t)^2\right)^{3/2}}{\left| x'(t) y''(t)-x''(t) y'(t)\right| }=\sqrt{x'(t)^2+y'(t)^2}
\end{align}
are
\begin{align}
t=2\pi n-\cos^{-1}(-5/4),n\in\mathbb{Z}\\
t=2\pi n+\cos^{-1}(-5/4),n\in\mathbb{Z}\\
t=\dfrac{2}{3}(3\pi n-\pi),n\in\mathbb{Z}\\
t=\dfrac{2}{3}(3\pi n+\pi),n\in\mathbb{Z}\\
\end{align}
as given by Wolfram|Alpha?

Version 1
For 
\begin{align}
&x(t)\text{:=}\cos (t)+\cos (2 t)+1&\\
&y(t)\text{:=}\sin (t)+\sin (2 t)&\\
&a(t)\text{:=}\sqrt{x'(t)^2+y'(t)^2}&\\
&b(t)\text{:=}\sqrt{x(t)^2+y(t)^2}&\\
\end{align}
how would I prove that the real roots of
$a(t)=\dfrac{2}{b(t)}$ are at $\pm\dfrac{2 \pi }{3}\pm2 \pi\  n,\  n\in \mathbb{Z}\ ?$
Update
Sorry for the lack of clarity. I am including a plot to help clarify what I mean:

As Git Gud pointed out in the comments, I think I am really after $$t\in\mathbb{R}:a(t)=\dfrac{2}{b(t)}$$
Update 2
I apologise to Galc127 who gave an excellent answer in spite of my mistake. $b$ should actually be
\begin{align}
b(t)\text{:=}\frac{\left| x'(t) y''(t)-x''(t) y'(t)\right| }{\left(x'(t)^2+y'(t)^2\right)^{3/2}}
\end{align}
This results in the plot given, and fits the solutions given - have been talking cross purposes as a result & am very grateful to Galc127 and others I have caused confusion to! Again, I apologise - feel free to downvote as seen fit!
NB Plot for original question:

So to clarify, question should have been what are solutions to
\begin{align}
t\in\mathbb{R}:\frac{2 \left(x'(t)^2+y'(t)^2\right)^{3/2}}{\left| x'(t) y''(t)-x''(t) y'(t)\right| }=\sqrt{x'(t)^2+y'(t)^2}
\end{align}
 A: Version I:
By simplifying:
$$x(t)=\cos(t)+\cos(2t)+1=\cos(t)+2\cos^2(t)-1+1=2\cos^2(t)+\cos(t) \\ x(t)=\cos(t)[2\cos(t)+1] \\ y(t):=\sin(t)+\sin(2t)=\sin(t)+2\sin(t)\cos(t) \\ y(t)=\sin(t)[2\cos(t)+1]$$
Therefore $x(t)=\cot(t)\cdot{y(t)}$.
Also:
$$x'(t)=-\sin(t)-2\sin(2t)=-y(t) \\ y'(t)=\cos(t)+2\cos(2t)$$
Now we can write: $$b(t)=\sqrt{x(t)^2+y(t)^2}=\sqrt{\cot^2(t)\cdot{y(t)^2}+y(t)^2}=\left|\frac{y(t)}{\sin(t)}\right|$$
If so, then $b(t)=\left|2\cos(t)+1\right|$.
Now, $$a(t)=\sqrt{x'(t)^2+y'(t)^2}=\sqrt{\left[-\left(\sin(t)+2\sin(2t)\right)\right]^2+\left[\cos(t)+2\cos(2t)\right]^2}$$
Simplifying:$$a(t)=\sqrt{\sin^2(t)+4\sin(t)\sin(2t)+4\sin^2(2t)+\cos^2(t)+4\cos(t)\cos(2t)+4\cos^2(2t)}$$
Using the identities $\displaystyle \sin^2x+\cos^2x=1$ and $\displaystyle \sin(x)\sin(y)+\cos(x)\cos(y)=\cos(x-y)$ we get $$a(t)=\sqrt{5+4\cos(t)}$$
Eventually, our equation becomes $$a(t)=\frac{2}{b(t)} \Rightarrow \sqrt{5+4\cos(t)}=\frac{2}{\left|2\cos(t)+1\right|}$$
Now, $5+4\cos(t)=2\left[2\cos(t)+1\right]+3$. Define $2\cos(t)+1=x$ we get the equation $$\sqrt{2x+3}=\frac{2}{|x|} \rightarrow |x|\sqrt{2x+3}=2 $$ 
This leads to a cubic equation which has one real root ($x \approx 0.91082$) and from here you can find the values of t.

Version II
$$x'(t)=-(\sin(t)+2\sin(2t)) \rightarrow x''(t)=-(\cos(t)+4\cos(2t)) \\ y'(t)=\cos(t)+2\cos(2t) \rightarrow y''(t)=-(\sin(t)+4\sin(2t))$$
Now we can find an expression for $b(t)$: $$\displaystyle b(t)=\frac{\left|x'(t)y''(t)-y'(t)x''(t)\right|}{\left(x'(t)^2+y'(t)^2\right)^{3/2}}\\ \small b(t)=\frac{\left|(\sin(t)+2\sin(2t))(\sin(t)+4\sin(2t))+(\cos(t)+2\cos(2t))(\cos(t)+4\cos(2t))\right|}{a(t)^3}$$
The equation is $a(t)=\frac{2}{b(t)}$, hence $$a(t)=\frac{2a(t)^3}{\left|(\sin(t)+2\sin(2t))(\sin(t)+4\sin(2t))+(\cos(t)+2\cos(2t))(\cos(t)+4\cos(2t))\right|}$$
We should simplify the denominator, so $$\displaystyle \small\left|(\sin(t)+2\sin(2t))(\sin(t)+4\sin(2t))+(\cos(t)+2\cos(2t))(\cos(t)+4\cos(2t))\right|= \\ =\displaystyle \small \left|\sin^2(t)+6\sin(t)\sin(2t)+\sin^2(2t)+\cos^2(2t)+6\cos(t)\cos(2t)+\cos^2(2t)\right|$$
Using the identities $\displaystyle \small \sin^2x+\cos^2x=1$ and $\displaystyle \small \sin(x)\sin(y)+\cos(x)\cos(y)=\cos(x-y)$ we get that $$\displaystyle b(t)=\frac{2a(t)^3}{\left|9+6\cos(t)\right|}$$
Our equation becomes $\displaystyle a(t)=\frac{2a(t)^3}{\left|9+6\cos(t)\right|}$.
It is clear that $a(t) \ne 0$, hence $\displaystyle \left|9+6\cos(t)\right|=2a(t)^2$.
From version I we know that $a(t)=\sqrt{5+4\cos(t)} \rightarrow a(t)^2=5+4\cos(t)$, thus $$\left|9+6\cos(t)\right|=2(5+4\cos(t))=10+8\cos(t)$$
There are two options to check in order to find the solutions.
