If y=y(x) satisfies the differential equation
$$16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos{y}dy=(1+2 \sin y)dx, x>0 \text{and} y(256) = \frac{\pi}{2}, y(49)=\alpha$$, then $$2\sin \alpha$$ is equal to :

$$16(\sqrt{\sqrt{x} + 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y dy = (1 + 2 \sin y)dx$$

$$\frac{\cos y}{1 + 2\sin y} dy = \frac{1}{16 \sqrt{x} \sqrt{9 + \sqrt{x}} (4 + \sqrt{9 + \sqrt{x}})} dx$$

• LHS: Let $$t = 1 + 2\sin y \implies dt = 2\cos y dy$$. Integral is $$\frac{1}{2} \ln(1 + 2\sin y)$$.
• RHS: Let $$u = 4 + \sqrt{9 + \sqrt{x}} \implies du = \frac{1}{4\sqrt{x}\sqrt{9 + \sqrt{x}}} dx$$. Integral is $$\frac{1}{4} \ln(4 + \sqrt{9 + \sqrt{x}})$$.

General Solution:

$$2 \ln(1 + 2\sin y) = \ln(4 + \sqrt{9 + \sqrt{x}}) + C \implies (1 + 2\sin y)^2 = K(4 + \sqrt{9 + \sqrt{x}})$$

$$(1 + 2\sin \frac{\pi}{2})^2 = K(4 + \sqrt{9 + \sqrt{256}})$$

$$3^2 = K(4 + \sqrt{25}) \implies 9 = 9K \implies K = 1$$

$$(1 + 2\sin \alpha)^2 = 4 + \sqrt{9 + \sqrt{49}}$$

$$2\sin \alpha = 2\sqrt{2} - 1$$