Can someone help me find ALL solutions in the interval [0,2π) for the given equations. Please show work.

Question

1) sin2x = sin2x

2) tan2x +tanx = 0

Answer 1

From my knowledge:
#=alpha
Sin:
x=#+2kpi and x=pi-#+2kpi

When u come out with the value of sine (alpha) you place it in the both cases and then you put it in the interval

0 <#+2kpi <2pi

Answer 2

#1 must be a joke. sin2x=sin2x for all x

#2

tan2x + tanx = 0

2tanx/(1-tan^2(x)) + tanx = 0
2tanx + tanx(1-tan^2(x)) = 0
tanx (2 + 1-tan^2x) = 0
tanx=0
tanx = ±√3

so, x=kπ or kπ ± π/3

Now just use the values of k so that the results come out in the desired domain.