The equation sin^2 theta - frac{4}{sin^3 thet | vedclass

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(D) Given the equation: $\sin^2 	heta - \frac{4}{\sin^3 	heta - 1} = 1 - \frac{4}{\sin^3 	heta - 1}$.Subtracting $-\frac{4}{\sin^3 	heta - 1}$ fro

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infinite roots

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(D) Given the equation: $\sin^2 	heta - \frac{4}{\sin^3 	heta - 1} = 1 - \frac{4}{\sin^3 	heta - 1}$.Subtracting $-\frac{4}{\sin^3 	heta - 1}$ from both sides,we get:$\sin^2 	heta = 1$.This implies $\sin 	heta = 1$ or $\sin 	heta = -1$.However,the term $\frac{4}{\sin^3 	heta - 1}$ is undefined when $\sin^3 	heta - 1 = 0$,which means $\sin 	heta = 1$.Therefore,$\sin 	heta = 1$ is not a valid solution.We are left with $\sin 	heta = -1$.The general solution for $\sin 	heta = -1$ is $	heta = 2n\pi - \frac{\pi}{2}$ for any integer $n$.Since there are infinitely many integers $n$,there are infinite roots.

The equation $\sin^2 \theta - \frac{4}{\sin^3 \theta - 1} = 1 - \frac{4}{\sin^3 \theta - 1}$ has:

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