For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$?

Something is amiss. We have

$$\int_0^{\pi/2} \frac{dx}{2+\sin (nx)} = \frac{1}{n}\int_0^{n\pi/2} \frac{dy}{2+\sin y}.$$

Using

$$\int_0^{2\pi} \frac{dy}{2+\sin y} = \frac{2\pi}{\sqrt{3}},$$

writing $n = 4k + r$ with $0 \leqslant r \leqslant 3$, we find

$$\int_0^{\pi/2}\frac{dx}{2+\sin (nx)} = \frac{k}{4k+r}\cdot \frac{2\pi}{\sqrt{3}} + \frac{1}{4k+r}\int_0^{r\pi/2}\frac{dy}{2+\sin y}.$$

In particular for $r = 0$, we have

$$\int_0^{\pi/2}\frac{dx}{2+\sin (4kx)} = \frac{1}{4}\frac{2\pi}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}} \neq \frac{\pi}{3\sqrt{3}}.$$

No $n$ divisible by $4$ satisfies the equation. For $n = 4k+1$, we have the value

$$\frac{k}{4k+1} \frac{2\pi}{\sqrt{3}} + \frac{1}{4k+1}\frac{\pi}{3\sqrt{3}} = \frac{6k+1}{4k+1} \frac{\pi}{3\sqrt{3}},$$

which only equals $\frac{\pi}{3\sqrt{3}}$ for $k = 0$.

For $n = 4k+2$, we have

$$\frac{k}{4k+2}\frac{2\pi}{\sqrt{3}} + \frac{2}{4k+2} \frac{\pi}{3\sqrt{3}} = \frac{6k+2}{4k+2}\frac{\pi}{3\sqrt{3}},$$

which also only equals $\frac{\pi}{3\sqrt{3}}$ for $k = 0$.

For $n = 4k+3$, we have

$$\frac{k}{4k+3}\frac{2\pi}{\sqrt{3}} + \frac{1}{4k+3}\frac{4\pi}{3\sqrt{3}} = \frac{6k+4}{4k+3}\frac{\pi}{3\sqrt{3}},$$

which never equals $\frac{\pi}{3\sqrt{3}}$. Thus $n = 1$ and $n = 2$ are the only solutions.

Sub $y=n x$ and use a Weierstrass substitution $t=\tan{(y/2)}$ and get that the integral is

$$\frac1{n} \int_0^{\tan{(\pi n/4)}} \frac{dt}{1+t+t^2} $$

Evaluate this by expressing the denominator as $(t+1/2)^2+3/4$ to get an expression for the integral and the stated equation:

$$\frac{2}{\sqrt{3} n} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{\pi}{4} n \right )+1}} \right ] } = \frac{\pi}{3 \sqrt{3}} $$

A quick scan shows this works for $n=1$ and $n=2$. I guess you would have to define the branch of the arctangent according to the value of $n$. For example, add $2 \pi$ to each multiple of $8$ for $n$. For example,

$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{\pi}{4} \right )+1}} \right ] } = \frac{\pi}{6} $$

$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{9 \pi}{4} \right )+1}} \right ] } = \frac{\pi}{6} + 2\pi$$

$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{17 \pi}{4} \right )+1}} \right ] } = \frac{\pi}{6} + 4\pi$$

and so on. Thus, for $n=12$, we have

$$\frac{2}{12 \sqrt{3}} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{12 \pi}{4} \right )+1}} \right ] } = \frac{1}{6 \sqrt{3}} 2 \pi = \frac{\pi}{3 \sqrt{3}} $$

For $n=13$, we have

$$\frac{2}{13\sqrt{3}} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{13 \pi}{4} \right )+1}} \right ] } = \frac{2 }{13 \sqrt{3}} \left ( 2 \pi+\frac{\pi}{6}\right ) = \frac{\pi}{3 \sqrt{3}} $$

Here is an alternative proof of the result, already posted by @Daniel Fisher, that the identity holds only for $n=1$ and $n=2$. My excuse to post this is that the approach below is simpler, actually it requires the value of none of these integrals.

Consider the integrals $$ a=\int_0^{\pi/2}\frac{\mathrm dx}{2+\sin x},\qquad b=\int_{-\pi/2}^0\frac{\mathrm dx}{2+\sin x}, $$ and note that, the sine function being positive on $(0,\frac\pi2)$ and negative on $(-\frac\pi2,0)$, $$ a\lt \frac\pi4\lt b. $$ The inequality $a\lt b$, and the periodicity of the sine function, are about everything we will need to know.

For every $n$, the change of variable $x\to nx$ shows that $$ \int_0^{\pi/2}\frac{\mathrm dx}{2+\sin(nx)}=\frac1n\sum_{k=1}^n\int_{(k-1)\pi/2}^{k\pi/2}\frac{\mathrm dx}{2+\sin x}. $$ Due to the periodicity of the sine function, the integrals in the sum on $k$ are alternatively $$ a,\quad a,\quad b,\quad b,\quad a,\quad a,\quad b,\quad b,\quad a,\ldots $$ hence the RHS, which is the barycenter of the $n$ first ones, is at least $a$ for every $n\geqslant1$ and equals $a$ if and only if no part equal to $b$ is involved, that is, for $n=1$ and $n=2$.

Although the problem has been solved a different method is presented here.

The problem asks to find which values of $n$ that satisfy \begin{align} \int_{0}^{\pi/2} \frac{dx}{2 + \sin(nx)} = \int_{0}^{\pi/2} \frac{dx}{2 + \sin(x)} = \frac{\pi}{3 \sqrt{3}}. \end{align}

The solution proposed is as follows.

Consider the integral \begin{align} I = \int_{0}^{\pi/2} \frac{dx}{a + \sin(nx)}. \end{align} The integral $I$ can be evaluated as \begin{align} I &= \left[ \frac{2}{ \sqrt{a^{2}-1} \ n } \tan^{-1} \left( \frac{a \tan\left(\frac{nx}{2}\right) + 1} { \sqrt{a^{2}-1}} \right) \right]_{0}^{\pi/2} \\ &= \frac{2}{\sqrt{a^{2}-1} \ n } \left[ \tan^{-1} \left( \frac{a \tan\left(\frac{n\pi}{4}\right) + 1} { \sqrt{a^{2}-1}} \right) - \tan^{-1} \left( \frac{1}{ \sqrt{a^{2}-1}} \right) \right] \\ &= \frac{2}{\sqrt{a^{2}-1} \ n } \tan^{-1} \left( \frac{ \sqrt{a^{2}-1} \ \theta_{n} }{ a + \theta_{n}} \right) \\ I &= \frac{2}{\sqrt{a^{2}-1} \ n } \tan^{-1} \left( \frac{ \sqrt{a^{2}-1} }{ a/\theta_{n} + 1} \right), \end{align} where $\theta_{n} = \tan(n\pi/4)$. Now, returning to the original integral, make the substitution $x = \pi/2 -u$ to obtain \begin{align} I &= \int_{0}^{\pi/2} \frac{du}{a + \sin(\frac{n \pi}{2} - nu)} \\ &= \int_{0}^{\pi/2} \frac{du}{a + \sin\left(\frac{n \pi}{2}\right) \cos(n u) - \cos\left( \frac{n \pi}{2}\right) \sin(n u)} \end{align} for which if the integral is to remain of the same form then it is required that \begin{align} \sin\left( \frac{n \pi}{2} \right) &= 0 \\ \cos\left( \frac{n \pi}{2} \right) &= -1. \end{align} From the sine equation it is seen that it is satisfied if $n = 2r$ for $r \geq 0$. The cosine equation is satisfied if $n = 4k+2$ for $k \geq 0$. It is evident that $2,6,10,\cdots$ is a subset of values in the set defined by $2r$ for $r \geq 0$. Since both equations are satisfied by the subset of values, this leads to the allowed values of $n$ for which the integral remains of the same form is $n = 4k+2$ for $k \geq 0$. This can be seen as \begin{align} \int_{0}^{\pi/2} \frac{dx}{a + \sin((4k+2)x)} = \int_{0}^{\pi/2} \frac{dx}{a + \sin(x)}, \end{align} for $k \geq 0$.

Now, with this $n$ value, it is seen that $\theta_{n}$ becomes \begin{align} \theta_{4k+2} &= \tan\left( \frac{(4k+2) \pi}{4} \right) = \tan\left( k \pi + \frac{\pi}{2} \right) \\ &= \frac{ \sin(k \pi + \pi/2) }{ \cos(k \pi + \pi/2) } = \frac{ (-1)^{k} }{ 0 } = \infty . \end{align} for which $1/\theta_{4k+2} = 0$. By using this result in the evaluation of the integral defined earlier provides \begin{align} I = \frac{2}{\sqrt{a^{2}-1} \ (4k+2) } \tan^{-1}( \sqrt{a^{2}-1} ). \end{align} Hence, \begin{align} \int_{0}^{\pi/2} \frac{dx}{a + \sin((4k+2)x)} &= \int_{0}^{\pi/2} \frac{dx}{a + \sin(x)} \\ &= \frac{1}{\sqrt{a^{2}-1} \ (2k+1) } \tan^{-1}( \sqrt{a^{2}-1} ), \end{align} for $k \geq 0$.

When $a=2$ the reduction leads to \begin{align} \int_{0}^{\pi/2} \frac{dx}{2 + \sin((4k+2)x)} &= \frac{ \tan^{-1}( \sqrt{3} ) }{\sqrt{3} \ (2k+1) } \\ &= \frac{ \tan^{-1}( \tan(\pi/3) ) }{\sqrt{3} \ (2k+1) } = \frac{\pi}{3 \sqrt{3} \ (2k+1) }. \end{align} It is now evident that the only value of $k$ that satisfies the desired result is $k=0$.

Hence, in terms of the original notation, the only values of $n$ are 1 and 2.