The sum of all values of $$x$$ in $$[0, 2\pi]$$, for which $$\sin x + \sin 2x + \sin 3x + \sin 4x = 0$$, is equal to:

We start from the given trigonometric equation

$$$\sin x+\sin 2x+\sin 3x+\sin 4x=0\;,\qquad x\in[0,\,2\pi].$$$

In order to combine the terms pair-wise, we recall the sum-to-product identity

$$\sin A+\sin B \;=\; 2\sin\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right).$$

Applying this once to the first and last terms and once to the two middle terms we obtain

$$$ \begin{aligned} \sin x+\sin 4x &= 2\sin\!\left(\dfrac{x+4x}{2}\right)\cos\!\left(\dfrac{x-4x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(-\dfrac{3x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(\dfrac{3x}{2}\right),\\[6pt] \sin 2x+\sin 3x &= 2\sin\!\left(\dfrac{2x+3x}{2}\right)\cos\!\left(\dfrac{2x-3x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(-\dfrac{x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(\dfrac{x}{2}\right). \end{aligned} $$$

Adding these two results we get

$$$ \sin x+\sin 2x+\sin 3x+\sin 4x = 2\sin\!\left(\dfrac{5x}{2}\right)\Bigl[\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)\Bigr]. $$$

Hence the original equation becomes

$$2\sin\!\left(\dfrac{5x}{2}\right)\Bigl[\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)\Bigr]=0.$$ So we must have

$$$\sin\!\left(\dfrac{5x}{2}\right)=0 \quad\text{or}\quad \cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)=0.$$$ We analyse the two cases separately.

Case 1. $$\sin\!\left(\dfrac{5x}{2}\right)=0.$$
The sine of any angle is zero when the angle equals an integer multiple of $$\pi$$, that is, when $$\dfrac{5x}{2}=n\pi,\qquad n\in\mathbb Z.$$ Solving for $$x$$ gives $$x=\dfrac{2n\pi}{5}.$$ Because $$x$$ must lie in $$[0,2\pi]$$ we list all permissible integers $$n$$:

$$$ \begin{aligned} n=0 &\;\Rightarrow\; x=0,\\ n=1 &\;\Rightarrow\; x=\dfrac{2\pi}{5},\\ n=2 &\;\Rightarrow\; x=\dfrac{4\pi}{5},\\ n=3 &\;\Rightarrow\; x=\dfrac{6\pi}{5},\\ n=4 &\;\Rightarrow\; x=\dfrac{8\pi}{5},\\ n=5 &\;\Rightarrow\; x=2\pi. \end{aligned} $$$

The sum of these six values equals

$$$ 0+\dfrac{2\pi}{5}+\dfrac{4\pi}{5}+\dfrac{6\pi}{5}+\dfrac{8\pi}{5}+2\pi =\dfrac{20\pi}{5}+2\pi =4\pi+2\pi =6\pi. $$$

Case 2. $$\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)=0.$$
For the cosine sum we employ the identity

$$\cos A+\cos B = 2\cos\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right).$$

Putting $$A=\dfrac{3x}{2}$$ and $$B=\dfrac{x}{2}$$ we get

$$\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right) =2\cos x\,\cos\!\left(\dfrac{x}{2}\right).$$

Thus the second factorised equation is

$$2\cos x\,\cos\!\left(\dfrac{x}{2}\right)=0,$$ which leads to

$$\cos x = 0 \quad\text{or}\quad \cos\!\left(\dfrac{x}{2}\right)=0.$$ We treat these two possibilities one by one.

(i) $$\cos x=0.$$
The cosine vanishes at odd multiples of $$\frac{\pi}{2}$$, so

$$x=\dfrac{\pi}{2}+n\pi,\qquad n\in\mathbb Z.$$

Inside the interval $$[0,2\pi]$$ this gives

$$x=\dfrac{\pi}{2},\; \dfrac{3\pi}{2}.$$

Their sum is

$$\dfrac{\pi}{2}+\dfrac{3\pi}{2}=2\pi.$$

(ii) $$\cos\!\left(\dfrac{x}{2}\right)=0.$$
Again, cosine is zero at odd multiples of $$\frac{\pi}{2}$$, so

$$\dfrac{x}{2}=\dfrac{\pi}{2}+m\pi,\qquad m\in\mathbb Z$$ which gives $$x=\pi+2m\pi.$$

Within $$[0,2\pi]$$ we obtain only

$$x=\pi,$$ whose contribution to the sum of roots is simply $$\pi.$$

Combining the results from both cases, we list all distinct solutions

$$$\Bigl\{\,0,\ \dfrac{2\pi}{5},\ \dfrac{4\pi}{5},\ \dfrac{6\pi}{5},\ \dfrac{8\pi}{5},\ 2\pi,\ \dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ \pi\Bigr\}.$$$ There are nine in total, and their sum equals

$$ 6\pi\;+\;2\pi\;+\;\pi =\;9\pi. $$

Hence, the correct answer is Option D.