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Question

(C) Given the differential equation: $\frac{dy}{dx} + \frac{1 + \cos 2y}{1 - \cos 2x} = 0$. Rearranging the terms,we get: $\frac{dy}{dx} = - \frac{1 +

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$	an y - \cot x = c$

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(C) Given the differential equation: $\frac{dy}{dx} + \frac{1 + \cos 2y}{1 - \cos 2x} = 0$. Rearranging the terms,we get: $\frac{dy}{dx} = - \frac{1 + \cos 2y}{1 - \cos 2x}$. Using the trigonometric identities $1 + \cos 2y = 2 \cos^2 y$ and $1 - \cos 2x = 2 \sin^2 x$,the equation becomes: $\frac{dy}{dx} = - \frac{2 \cos^2 y}{2 \sin^2 x} = - \frac{\cos^2 y}{\sin^2 x}$. Separating the variables,we have: $\frac{dy}{\cos^2 y} = - \frac{dx}{\sin^2 x}$,which is equivalent to $\sec^2 y \, dy = - \csc^2 x \, dx$. Integrating both sides: $\int \sec^2 y \, dy = - \int \csc^2 x \, dx$. This yields: $	an y = -(-\cot x) + c$,which simplifies to $	an y = \cot x + c$. Thus,the solution is $	an y - \cot x = c$.

The solution of the differential equation $\frac{dy}{dx} + \frac{1 + \cos 2y}{1 - \cos 2x} = 0$ is:

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