The sum of all values of x in [ 0, 2π ] , for which x + 2x + 3x +...

Given,

$\sin x + \sin 2x + \sin 3x + \sin 4x = 0$

$⇒ ( \sin x + \sin 4x ) + ( \sin 2x + \sin 3x ) = 0$

$⇒ 2sin\dfrac{5x}{2}cos\dfrac{3x}{2} + 2sin\dfrac{5x}{2}cos\dfrac{x}{2} = 0$

$⇒ 2sin\dfrac{5x}{2}\left\{cos\dfrac{3x}{2} + cos\dfrac{x}{2}\right\} = 0$

$⇒ 2sin\dfrac{5x}{2}\left\{2\cos x \cos\dfrac{x}{2}\right\} = 0$

$2sin\dfrac{5x}{2} = 0 ⇒ \dfrac{5x}{2} = 0, \pi, 2\pi, 3\pi, 4\pi, 5\pi$

$⇒ x = 0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}, \dfrac{6\pi}{5}, \dfrac{8\pi}{5}, 2\pi$

and $cos\dfrac{x}{2} = 0 ⇒ \dfrac{x}{2} = \dfrac{\pi}{2} ⇒ x = \pi$

and $\cos x = 0 ⇒ x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$

So sum $= 6\pi + \pi + 2\pi = 9\pi$