An unusual limit of lengths of intervals

Let's consider what is going on when n is already large, this way we will get the idea towards the solution.

If n is big, $\displaystyle \sin(4nx)$ oscillates very frequently and 1 interval is $\displaystyle \dfrac{\pi}{2n}$ which is very small in the limit n tends to infinity. More precisely, the variation of $\displaystyle \sin(x)$ function is of order $\displaystyle \dfrac{1}{n}$ across this 1 cycle interval. So, we can regard this as a constant.

Now, if A is a positive constant, the length of interval of $\displaystyle \sin(x)$ such that the function lies above the level A is $\displaystyle \pi-2\arcsin(A)$

In our case, we split $\displaystyle [0,\pi/2]$ into n intervals of equal length equals to $\displaystyle \dfrac{\pi}{2n}$ , so, the sum of all intervals satisfy the condition is then $\displaystyle \Sigma_{m=0}^{n-1} \dfrac{1}{4n}(\pi-2\arcsin(\sin(\dfrac{m\pi}{2n})) )$ .

We can sum these explicitly, we get $\displaystyle \dfrac{\pi}{4}( 1-\dfrac{n+1}{2n})$ in the limit n tends to infinity, we get $\displaystyle S_{n}=\dfrac{\pi}{8}$ , so $1+8=\boxed{9}$