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(C) Given $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$.Since both $\sin x$ and $\cos x$ are negative,$x$ lies in the third quadrant.We know tha

Vedclass · Accepted Answer

$x = n\pi + 	an^{-1}\left(\frac{3}{4}ight), n \in Z$

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(C) Given $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$.Since both $\sin x$ and $\cos x$ are negative,$x$ lies in the third quadrant.We know that $	an x = \frac{\sin x}{\cos x} = \frac{-3/5}{-4/5} = \frac{3}{4}$.The general solution for $	an x = 	an \alpha$ is $x = n\pi + \alpha$,where $n \in Z$.Here,$	an x = \frac{3}{4}$,so $x = n\pi + 	an^{-1}\left(\frac{3}{4}ight)$.Since $x$ must be in the third quadrant,we choose $n$ such that $x$ satisfies the signs of $\sin x$ and $\cos x$. Specifically,for $n$ being an odd integer,$x$ falls in the third quadrant.Thus,the general solution is $x = n\pi + 	an^{-1}\left(\frac{3}{4}ight)$ for $n \in Z$.

The general solution satisfying both the equations $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$ is

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