Question

Find all solutions in the interval [0, 2π).

sin2x + sin x = 0

Answer 1

can someone please help!:(

Answer 2

sin2x=- sinx
sin 2x= sin(-x)
2x=-x +2k pi
2x=180-(-x) +2k pi

Answer 3

2x=-x+2k pi --->3x=2k pi
x=2k pi /3 so x=0,2pi/3 ,4pi/3

and 2x=180+x+2k pi ----> x=180 +2k pi so just x=180


total roots in your interval are 0, 2pi/3, 4pi/3,pi

Answer 4

@amoodarya what about pi/3?

Answer 5

sin(2)(pi/3) +sinpi/3 =sin 120 +sin 60 =sqrt3/2+sqrt3/2 =sqrt3
so pi/3 is not a root

Answer 6

sin 2x=2 sinx cosx
so
other solving
2 sinx cos x +sinx =0
sinx(2cos x+1)=0
so sinx =0
and cos x=-1/2

Answer 7

a. x = 0, π, 4pi/3
b. x = 0, π,pi/3,2pi/3
c. x = 0, π, pi/3,5pi/3
d. x = 0, π, 3pi/2

Answer 8

These are the option i have..