How would you solve this equation for b in terms of c?

Question

How would you solve this equation for b in terms of c?

1700b(sin(14b/3 + c)) + 68cos(14b/3 + c) - (10625b^2)(cos(14b/3 + c) = 0

Answer 1

Let w = 14b/3 + c, and rearrange as follows: (10625 cos w) b² - (1700 sin w) b - 68 cos w = 0.

Quadratic formula yields b = {[1700 sin w ± √((-1700 sin w)² - 4(10625 cos w)(-68 cos w))] / (2•10625 cos w)}.

Under the square root we have 2,890,000 sin² w + 2,890,000 cos² w = 2,890,000 (sin² w + cos² w = 1), and √2,890,000 = 1700.

Then b = 0.08 tan (14b/3 + c) ± 0.08 sec(14b/3 + c). [since tan w = (sin w)/(cos w) and sec w = 1/cos w]