How do you solve #Sin x + Sin 2x + Sin 3x + Sin 4x = 0#?

In this way, using the sum-to-product formula for sinus and cosine:

#sinalpha+sinbeta=2sin((alpha+beta)/2)cos((alpha-beta)/2)#

#cosalpha+cosbeta=2cos((alpha+beta)/2)cos((alpha-beta)/2)#.

So:

#(sin4x+sinx)+(sin3x+sin2x)=0rArr#

#2sin((4x+x)/2)cos((4x-x)/2)+#

#+2sin((3x+2x)/2)cos((3x-2x)/2)=0#

#2sin(5/2x)cos(3/2x)+2sin(5/2x)cos(1/2x)=0#

#2sin(5/2x)[cos(3/2x)+cos(1/2x)]=0#

#2sin(5/2x)2cos((3/2x+1/2x)/2)cos((3/2x-1/2x)/2)=0#

#4sin(5/2x)cosxcos(x/2)=0#.

Then:

#sin(5/2x)=0rArr5/2x=kpirArrx=2/5kpi#,

#cosx=0rArrx=pi/2+kpi#,

#cos(x/2)=0rArrx/2=pi/2+kpirArrx=pi+2kpi#.

Reminder of Concept to solve trig equations:
To solve a complex trig equation, transform it into a few basic trig equations. Solving trig equations finally results in solving basic trig equations.
In this example of solving: f(x) = sin x + sin 2x + sin 3x + sin 4x = 0, use the trig identity of "sin a + sin b" to transform f(x) into a product of 3 basic trig equations like Massimiliano finely did:
#f(x) =4cos x.sin (5x/2).cos (x/2) = 0#.
Next, solve separately the 3 basic trig equations: cos x = 0, sin #(5x/2) = 0, and cos (x/2) = 0#