The general solution of frac{1-cos 2x}{1+cos | vedclass

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(D) Given the equation $\frac{1-\cos 2x}{1+\cos 2x}=3$. Using the trigonometric identities $1-\cos 2x = 2\sin^2 x$ and $1+\cos 2x = 2\cos^2 x$,we get:

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$x=n\pi \pm \frac{\pi}{3}, n \in Z$

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(D) Given the equation $\frac{1-\cos 2x}{1+\cos 2x}=3$. Using the trigonometric identities $1-\cos 2x = 2\sin^2 x$ and $1+\cos 2x = 2\cos^2 x$,we get: $\frac{2\sin^2 x}{2\cos^2 x} = 3$ $	an^2 x = 3$ Since $	an^2 x = 3$,we have $	an^2 x = (\sqrt{3})^2 = 	an^2 \frac{\pi}{3}$. The general solution for $	an^2 x = 	an^2 \alpha$ is $x = n\pi \pm \alpha$. Therefore,$x = n\pi \pm \frac{\pi}{3}, n \in Z$.

The general solution of $\frac{1-\cos 2x}{1+\cos 2x}=3$ is

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