Use the double angle identity:
sin(2x) = 2 sin(x) cos(x)
Now rewrite
sin(2x) sin(x) + cos(x) = 0
as
2 sin²(x) cos(x) + cos(x) = 0
Factor out cos(x) :
cos(x) (2 sin²(x) + 1) = 0
Consider the two cases,
cos(x) = 0 OR 2 sin²(x) + 1 = 0
Solve for cos(x) and sin²(x) :
cos(x) = 0 OR sin²(x) = -1/2
Squaring a real number always gives a non-negative number, so the second case doesn't offer any real solutions. We're left with
cos(x) = 0
Cosine is zero for odd multiples of π/2, so we have
x = (2n + 1) π/2
where n is any integer.