From the formula for #sin (2theta)# we have
#sin 2theta = 2sin theta cos theta#
From Pythagoras theorem we have
#sin^2 theta + cos^2 theta = 1#
So
#sin 2theta sin theta#
#= (2sin theta cos theta)sin theta#
#= 2sin^2 theta cos theta#
#=2(1-cos^2 theta)cos theta#
Putting this together with your equation, we get
#2(1-cos^2 theta)cos theta = cos theta#
If #cos theta = 0# then both sides will be zero.
So some solutions to the original problem are:
#theta = pi/2 + npi# for all #n# in #ZZ#.
On the other hand, if #cos theta != 0#, divide both sides of the equation by #cos theta# to get
#2(1-cos^2 theta) = 1#
Divide both sides by 2 to get
#1-cos^2 theta = 1/2#
So #cos^2 theta = 1/2# and #cos theta = +-1/sqrt(2)#
This is true for
#theta = pi/4 + (npi)/2# for all #n# in #ZZ#.
