# STEP III 2003 Q6

For the second part of this question I only got one set of solutions to the equation which was : θ=((2n+1)pi)/(b+a-1) however I did not get the second set of solutions that were given in the official solution θ=(2npi)/(b-a).
My solution is roughly the same as the solution in the STEP III algebra module Q2 however after I got (sin(b-1/2)θ - sin(a-1/2)θ)/(sin(θ/2))=0, I expanded the two sine terms on the numerator using the compound angle formula and divided by sin(θ/2) on top and bottom of the fraction. Then I got cot(θ/2)=(cosbθ - cosaθ)/(sinbθ -sinaθ). I then used the sum to product formula on the RHS and got tan(θ/2)=-cot((a+b)θ/2). Then using the relation between the cot and tan graphs and the general solution formula for tan which yielded θ=((2n+1)pi)/(1-b-a).

Can anyone tell me where the second set of solutions went missing in my answer?

### I think the solution went

I think the solution went missing in this step: $$(\sin b\theta - \sin a\theta)\cot \tfrac{1}{2}\theta=\cos b\theta-\cos a\theta \quad \overset{?}{\implies} \quad\cot \tfrac{1}{2}\theta=\frac{\cos b\theta-\cos a\theta}{\sin b\theta - \sin a\theta},$$ since your missing set of solutions, $\theta=2n\pi/(b-a),$ satisfy $\sin b\theta - \sin a\theta=0$.

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