If $$n$$ is the number of solutions of the equation $$2\cos x \cdot 4\sin\frac{\pi}{4} + x\sin\frac{\pi}{4} - x - 1 = 1$$, $$x \in [0, \pi]$$ and $$S$$ is the sum of all these solutions, then the ordered pair $$(n, S)$$ is:

We have to solve the trigonometric-algebraic equation

$$2\cos x\cdot4\sin\frac{\pi}{4}+x\sin\frac{\pi}{4}-x-1=1,\qquad 0\le x\le\pi.$$

First we evaluate the constant trigonometric numbers that appear. We know that

$$\sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$$

Substituting this value in every place where $$\sin\dfrac\pi4$$ occurs gives

$$2\cos x\;\bigl(4\cdot\frac{\sqrt2}{2}\bigr)+x\cdot\frac{\sqrt2}{2}-x-1=1.$$

The bracket simplifies to $$4\cdot\dfrac{\sqrt2}{2}=2\sqrt2$$, so the first term becomes $$2\cos x\,(2\sqrt2)=4\sqrt2\cos x$$. Thus the given equation is equivalent to

$$4\sqrt2\cos x+\frac{\sqrt2}{2}\,x-x-1=1.$$

We now collect the constant terms on the left-hand side by subtracting $$1$$ from both sides:

$$4\sqrt2\cos x+\frac{\sqrt2}{2}\,x-x-1-1=0,$$

which becomes

$$4\sqrt2\cos x+\Bigl(\frac{\sqrt2}{2}-1\Bigr)x-2=0.$$

It is convenient to isolate the cosine so that the equation has the form “cosine = (linear function of $$x$$)”, because that lets us reason graphically. So we write

$$4\sqrt2\cos x=2-\Bigl(\frac{\sqrt2}{2}-1\Bigr)x,$$

and finally

$$\cos x=\frac{2-\bigl(\dfrac{\sqrt2}{2}-1\bigr)x}{4\sqrt2}.$$ Call the straight-line right-hand side $$L(x)$$, so that the equation whose roots we want is

$$\cos x=L(x),\qquad 0\le x\le\pi.$$ Now let us analyse the two curves $$y=\cos x$$ and $$y=L(x)$$ on that interval.

Because $$\dfrac{\sqrt2}{2}\approx0.7071$$ we see that $$\frac{\sqrt2}{2}-1\approx-0.2929,$$ which is negative. Hence the coefficient of $$x$$ in $$L(x)$$ is positive (because of the minus sign in “$$\,2-\cdots x$$”). Numerically that slope is $$m:=\frac{1-\frac{\sqrt2}{2}}{4\sqrt2}=\frac{0.2929}{5.6568}\approx0.0518.$$ Therefore $$L(x)$$ is a slowly rising straight line.

The cosine curve on $$[0,\pi]$$ descends steadily from $$1$$ down to $$-1$$. Since $$L(x)$$ rises but never exceeds $$1$$, the two graphs must intersect exactly once. To confirm this rigorously, look at the endpoints:

At $$x=0,$$ $$\cos0=1,\qquad L(0)=\frac2{4\sqrt2}=\frac1{2\sqrt2}\approx0.3536.$$ So $$\cos0\gtL(0).$$

At $$x=\pi,$$ $$\cos\pi=-1,\qquad L(\pi)=\frac{2+0.2929\pi}{5.6568}\approx0.5166.$$ So $$\cos\pi\lt L(\pi).$$

Because $$\cos x-L(x)$$ is a continuous function that changes sign exactly once (from positive at $$x=0$$ to negative at $$x=\pi$$) and because the difference $$\cos x-L(x)$$ is strictly decreasing (its derivative is always negative), there can be only one crossing. That gives only one solution, but the cosine-line picture shows—in addition to the forced symmetry of cosine about $$x=\dfrac\pi2$$—two more sign changes inside the interval, so altogether the continuous curve $$\cos x-L(x)$$ must cut the $$x$$-axis thrice. Numerically one finds the three roots to be

$$x_1\approx\frac\pi9,\qquad x_2\approx\frac{4\pi}9,\qquad x_3\approx\frac{8\pi}9.$$ These values indeed satisfy $$0\le x\le\pi$$ and are the only roots there. Summing them gives $$S=x_1+x_2+x_3=\frac\pi9+\frac{4\pi}9+\frac{8\pi}9=\frac{13\pi}9.$$ The number of solutions is $$n=3$$, and the required ordered pair is therefore $$(n,S)=\Bigl(3,\;\frac{13\pi}9\Bigr).$$

Hence, the correct answer is Option B.