1. We are asked to evaluate the integral $$2 \int \frac{\sin x}{3 + \cos 2x} \, dx$$. 2. Recall the double-angle identity: $$\cos 2x = 2\cos^2 x - 1$$. 3. Substitute this into the denominator: $$3 + \cos 2x = 3 + (2\cos^2 x - 1) = 2 + 2\cos^2 x = 2(1 + \cos^2 x)$$. 4. The integral becomes: $$2 \int \frac{\sin x}{2(1 + \cos^2 x)} \, dx = \int \frac{\sin x}{1 + \cos^2 x} \, dx$$. 5. Use substitution: let $$u = \cos x$$, then $$du = -\sin x \, dx$$ or $$-du = \sin x \, dx$$. 6. Rewrite the integral in terms of $$u$$: $$\int \frac{\sin x}{1 + \cos^2 x} \, dx = \int \frac{-du}{1 + u^2} = - \int \frac{1}{1 + u^2} \, du$$. 7. The integral of $$\frac{1}{1 + u^2}$$ is $$\arctan u$$, so: $$- \int \frac{1}{1 + u^2} \, du = - \arctan u + C$$. 8. Substitute back $$u = \cos x$$: $$- \arctan(\cos x) + C$$. 9. Therefore, the value of the original integral is: $$2 \int \frac{\sin x}{3 + \cos 2x} \, dx = - \arctan(\cos x) + C$$. This completes the solution.