If f(x) = frac{log_{sin |x|} cos^3 x}{log_{si | vedclass

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(C) The function is defined as $f(x) = \frac{3 \log_{\sin |x|} \cos x}{3 \log_{\sin |3x|} \cos (x/2)} = \frac{\ln \cos x}{\ln \sin |x|} 	imes \frac{\

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$2$

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(C) The function is defined as $f(x) = \frac{3 \log_{\sin |x|} \cos x}{3 \log_{\sin |3x|} \cos (x/2)} = \frac{\ln \cos x}{\ln \sin |x|} 	imes \frac{\ln \sin |3x|}{\ln \cos (x/2)}$.For $f(x)$ to be defined,the base of the logarithm must be positive and not equal to $1$,and the argument must be positive.$1$. $\sin |x| > 0$ and $\sin |x| 
eq 1 \implies |x| \in (0, \pi) \setminus \{\pi/2\}$. Given $|x| $2$. $\sin |3x| > 0$ and $\sin |3x| 
eq 1 \implies |3x| \in (0, \pi) \setminus \{\pi/2\} \implies |x| \in (0, \pi/3) \setminus \{\pi/6\}$.Thus,$f(x)$ is undefined at $x = \pm \pi/6$ and $x = 0$ (initially). Since $f(0)=4$,we check $\lim_{x 	o 0} f(x)$.Using $\ln \cos x \approx -x^2/2$,$\ln \sin |x| \approx \ln |x|$,$\ln \sin |3x| \approx \ln |3x|$,$\ln \cos (x/2) \approx -x^2/8$,we get $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{-x^2/2}{\ln |x|} \cdot \frac{\ln |3x|}{-x^2/8} = 4 \lim_{x 	o 0} \frac{\ln |3x|}{\ln |x|} = 4 \lim_{x 	o 0} \frac{\ln 3 + \ln |x|}{\ln |x|} = 4(1) = 4$.Since $\lim_{x 	o 0} f(x) = f(0) = 4$,$f$ is continuous at $x = 0$.The points of discontinuity are $x = \pi/6$ and $x = -\pi/6$. Total $2$ points.

If $f(x) = \frac{\log_{\sin |x|} \cos^3 x}{\log_{\sin |3x|} \cos^3 (x/2)}$ for $|x| < \frac{\pi}{3}, x \neq 0$ and $f(0) = 4$,then the number of points of discontinuity of $f$ in $\left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ is:

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