<s>[INST] Finding the sum of the real roots of $(\tan(x)-\sqrt{3})\cdot \arcsin \dfrac{2(x-\pi)}{\pi}=0$ 
Find the sum of the real roots of
$$(\tan(x)-\sqrt{3})\cdot \arcsin \dfrac{2(x-\pi)}{\pi}=0$$

can anybody assist with my solution.

*

*$$\tan(x)=\sqrt{3}$$
$$x=\frac{\pi}{3}$$ and $$x=\frac{4\pi}{3}$$. On the other hand $$x=\frac{\pi}{3}+ \frac{k}{\pi}$$, where $k∈\mathbb Z$.


*$$\arcsin(\frac{2(x-\pi)}{\pi})=0\implies 
x=\pi$$
Answer: $$\frac{\pi}{3}+\pi=\frac{4\pi}{3} $$(got negative result with this. No instructions about $k∈\mathbb Z$)
$$\pi+\frac{4\pi}{3}+\pi=\frac{8\pi}{3} \tag{?}$$
Thank you for help!
 [/INST] We need to ensure that the roots lie within the domain of definition of arcsin.
That is, we need
$$-1 \leq \frac{2(x-\pi)}{\pi} \leq 1.$$
We know that $x = \pi$ is a solution that makes this equal to zero.  But, it can also be the case that the tangent evaluates to zero, but not the arcsin.  We substitute $x = \pi/3+k\pi$ into the above equation.  This gives (after simplification),
$$\frac{1}{6} \leq k \leq \frac{7}{6},$$
from which we conclude that $k$ can only be $1$.
Thus, the roots are $\pi$ and $\pi/3+\pi$.
The sum is then $2\pi+\pi/3 = 7\pi/3$.
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