1. **Stating the problem:** Simplify the expression $$\frac{3\sin 2x + 5\cos x}{5\sin 2x - 2\cos 2x}$$ and analyze the relationship with $$\tan x$$. 2. **Recall formulas:** - Double angle formulas: $$\sin 2x = 2\sin x \cos x$$ $$\cos 2x = \cos^2 x - \sin^2 x$$ 3. **Rewrite numerator and denominator using double angle formulas:** Numerator: $$3\sin 2x + 5\cos x = 3(2\sin x \cos x) + 5\cos x = 6\sin x \cos x + 5\cos x$$ Denominator: $$5\sin 2x - 2\cos 2x = 5(2\sin x \cos x) - 2(\cos^2 x - \sin^2 x) = 10\sin x \cos x - 2\cos^2 x + 2\sin^2 x$$ 4. **Factor where possible:** Numerator: $$\cos x (6\sin x + 5)$$ Denominator: Rewrite denominator as: $$10\sin x \cos x - 2\cos^2 x + 2\sin^2 x = 2\sin^2 x + 10\sin x \cos x - 2\cos^2 x$$ 5. **Express denominator in terms of $\sin x$ and $\cos x$:** Group terms: $$2\sin^2 x - 2\cos^2 x + 10\sin x \cos x = 2(\sin^2 x - \cos^2 x) + 10\sin x \cos x$$ 6. **Recall that $\sin^2 x - \cos^2 x = -(\cos 2x)$, but better to keep as is for now.** 7. **Final simplified form:** $$\frac{\cos x (6\sin x + 5)}{2(\sin^2 x - \cos^2 x) + 10\sin x \cos x}$$ 8. **Relation to $\tan x$:** Since $\tan x = \frac{\sin x}{\cos x}$, the expression involves $\sin x$ and $\cos x$ terms but is not a simple reciprocal or direct function of $\tan x$ without further manipulation. **Answer:** $$\boxed{\frac{3\sin 2x + 5\cos x}{5\sin 2x - 2\cos 2x} = \frac{\cos x (6\sin x + 5)}{2(\sin^2 x - \cos^2 x) + 10\sin x \cos x}}$$