determinati solutiile ecuatiei sinx+sin2x+sin3x=1+cosx+cos2x, care apartine intervalului (pi/2;5pi/6)

sin3x+sinx+sin2x=1+cos2x+cosx
2sin2x*cosx+sin2x=2(cosx)^2+cosx
sin2x*(2cosx+1)-cosx(2cosx+1)=0
2sinxcosx(2cosx+1)-cosx(2cosx+1)=0
cosx(2cosx+1)(2sinx-1)=0
cox=0=>x=pi/2 +k*pi , k apartine Z =>x=pi/2
2cosx+1=0 =>cox=-1/2 => x apartine {2pi/3 +2k*pi|k intreg}U{4pi/3 +2k*pi|k intreg} =>x=2pi/3
2sinx-1=0 => x apartine {pi/6 +2k*pi|k intreg}U{5pi/6+2k*pi|k intreg} => x=5pi/6

daca e interval deschis x=2pi/3

[Citat] determinati solutiile ecuatiei sinx+sin2x+sin3x=1+cosx+cos2x, care apartine intervalului (pi/2;5pi/6)

va multumesc

[Citat] sin3x+sinx+sin2x=1+cos2x+cosx 2sin2x*cosx+sin2x=2(cosx)^2+cosx sin2x*(2cosx+1)-cosx(2cosx+1)=0 2sinxcosx(2cosx+1)-cosx(2cosx+1)=0 cosx(2cosx+1)(2sinx-1)=0 cox=0=>x=pi/2 +k*pi , k apartine Z =>x=pi/2 2cosx+1=0 =>cox=-1/2 => x apartine {2pi/3 +2k*pi|k intreg}U{4pi/3 +2k*pi|k intreg} =>x=2pi/3 2sinx-1=0 => x apartine {pi/6 +2k*pi|k intreg}U{5pi/6+2k*pi|k intreg} => x=5pi/6 daca e interval deschis x=2pi/3

sin3x+sinx este egal cu :2sin2x*cosx?