# Show equation has at most two solutions on (0,2*Pi)

$\sin (\text{ha})\text{ = } \text{dec}'(\text{ha}) (\tan (\text{lat})-\cos (\text{ha}) \tan (\text{dec}(\text{ha})))$

I want to show this equation has at most two solutions for 0 < ha < 2*Pi under these conditions, for ANY function dec[ha] and ANY value of lat satisfying the conditions below:

• lat is fixed and between -Pi/2 and Pi/2, noninclusive

• Abs[dec[ha]] is bounded by Pi/6

• Abs[dec'[ha]] is bounded by 1/36, and dec[ha] is everywhere differentiable.

If this can be shown, what is the weakest bound I can put on dec'[ha] so that this equation still has at most 2 solutions in (0,2*Pi)?

Purpose: resolve http://astronomy.stackexchange.com/questions/962/

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it might be in your link, but I think you owe it to us to define what "dec" means in your question. This is a math forum, and I don't think that's a standard math term. – Stefan Smith Nov 30 '13 at 18:29
Thanks, edited. I want this for ANY function dec[ha] (that satisfies the given conditions). – barrycarter Nov 30 '13 at 18:30