Sin x + sin 3x = sin 4x

[latex]sinx+sin3x=sin4x \\ sinx+sin3x=2sin2x*cos2x \\ 2sin \frac{x+3x}{2} *cos \frac{x-3x}{2}=2sinxcosx*(1-2 sin^{2}x) \\ 2sin2x*cos(-x)-sin2x(1-2cos^{2} x-1)=0 \\ sin2x(2cos(-x)+2cos^{2}x+1-1)=0 \\ 1) sin2x=0 \\ 2x= \pi n,n e Z \\ x= \frac{ \pi }{2}n, neZ \\ ******* \\ 2cos(-x)+cos2x+1=0 \\ 4cos \frac{-x+2x}{2}*cos \frac{-x-2x}{2}-1=0 \\ 4cos \frac{x}{2}*cos \frac{-3x}{2}=1 \\ 2) 4cos \frac{x}{2}=1 \\ cos \frac{x}{2}= \frac{1}{4} \\ \frac{x}{2}=arccos \frac{1}{4} +n,neZ \\ x= \frac{ \pi }{3} +2 \pi n,neZ[/latex] [latex]3) cos \frac{-3}{2}x=1 \\ \frac{-3}{2}x=2 \pi n,neZ (* \frac{-2}{3}) \\ x= \frac{-4}{3} \pi n,neZ [/latex]