The number of solutions of the equation sin l | vedclass

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(B) Let $	heta = 2 	an^{-1} x$. Then $\cot 	heta = \cot(2 	an^{-1} x) = \frac{1 - x^2}{2x}$.Given equation is $\sin(2 \cos^{-1}(\cot 	heta)) = 0$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $	heta = 2 	an^{-1} x$. Then $\cot 	heta = \cot(2 	an^{-1} x) = \frac{1 - x^2}{2x}$.Given equation is $\sin(2 \cos^{-1}(\cot 	heta)) = 0$.This implies $2 \cos^{-1}(\cot 	heta) = n\pi$ for some integer $n$.So,$\cos^{-1}(\cot 	heta) = \frac{n\pi}{2}$,which means $\cot 	heta = \cos(\frac{n\pi}{2})$.For the range of $\cos^{-1}$,we have $0 \le \cos^{-1}(\cot 	heta) \le \pi$,so $n$ can be $0, 1, 2$.Case $1$: $n=0 \implies \cos^{-1}(\cot 	heta) = 0 \implies \cot 	heta = 1 \implies \frac{1 - x^2}{2x} = 1 \implies x^2 + 2x - 1 = 0 \implies x = -1 \pm \sqrt{2}$.Case $2$: $n=1 \implies \cos^{-1}(\cot 	heta) = \frac{\pi}{2} \implies \cot 	heta = 0 \implies \frac{1 - x^2}{2x} = 0 \implies x^2 = 1 \implies x = \pm 1$.Case $3$: $n=2 \implies \cos^{-1}(\cot 	heta) = \pi \implies \cot 	heta = -1 \implies \frac{1 - x^2}{2x} = -1 \implies x^2 - 2x - 1 = 0 \implies x = 1 \pm \sqrt{2}$.All these $6$ values are valid as they are in the domain of $	an^{-1} x$ and satisfy the range of $\cos^{-1}$.Thus,there are $6$ solutions.

The number of solutions of the equation $\sin \left[2 \cos^{-1} \left\{\cot \left(2 \tan^{-1} x\right)\right\}\right] = 0$ is

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