How do you solve #sin(2theta)+sintheta=0# on the interval #0<=theta<2pi#?

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Answer 1 · 2015-04-27T18:22:30

#theta=0, {2pi}/3, pi or {4pi}/3#

Use #sin(2 theta) = 2 sintheta costheta#

Then factor and solve.

#sin(2 theta) + sintheta =0#

# 2 sintheta costheta + sintheta =0#

#sin theta (2cos theta +1) =0#

#sin theta = 0##color(white)"sssss"# or #color(white)"sssss"# #2 cos theta + 1 =0 # so

#sin theta = 0##color(white)"sssss"# or #color(white)"sssss"# #cos theta = -1/2#

#theta=0, {2pi}/3, pi or {4pi}/3#