Compound angles help

I can't figure this compound angles question out for the life of me.
Question:
Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ.

Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).

You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.

Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?

Thanks for the pointer. Using Sin (A+B) i've changed the equation to:
Sin(2θ+θ) = Sinθ and from there using the rule,
Sin2θcosθ + Cos2θsinθ = sinθ

However, I've hit a wall again as to where to go from there.

You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.

Thank you! I never even considered the Identities.

Substituting the identities in:

(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get

Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π

Thank you! You're a life saver!

Thank you! I never even considered the Identities.

Substituting the identities in:

(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get

Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π

Thank you! You're a life saver!

Following on from Notnek's suggestion you could look at:

sin 3θ - sin θ = 0 then use the formula to turn this into a product