Express the equation cosecθ(3 cos 2θ+7)+11=0 in the form asin^2(θ) + bsin(θ) + c = 0, where a, b and c are constants.

We must first use the identity cosecθ = 1/sinθ. Now the equation becomes (1/sinθ)(3 cos 2θ+7)+11=0. Since we know that the question is asking for the answer in the form of asin²θ + bsinθ + c = 0, we realise that we must change cos 2θ into something containing sinθ instead. To do this we use the double angle formula cos2θ=1-2sin²θ. Substituting this into the equation we get (1/sinθ)(3 (1-2sin²θ)+7)+11=0. Simplifying this by multiplying out the brackets we find; (1/sinθ)(10-6sin²θ)+11=0. Multiplying out again gives us (10/sinθ)-6sinθ+11=0. Comparing this with the form of the answer that the question requires means we must multiply through by sinθ, giving us -6sin²θ + 11sinθ + 10 = 0.