The number of solutions of tan(5pi cos theta) | vedclass

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(A) Given $	an(5\pi \cos 	heta) = \cot(5\pi \sin 	heta)$.Using the identity $\cot(x) = 	an(\frac{\pi}{2} - x)$,we have $	an(5\pi \cos 	heta) = \

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) Given $	an(5\pi \cos 	heta) = \cot(5\pi \sin 	heta)$.Using the identity $\cot(x) = 	an(\frac{\pi}{2} - x)$,we have $	an(5\pi \cos 	heta) = 	an(\frac{\pi}{2} - 5\pi \sin 	heta)$.The general solution is $5\pi \cos 	heta = n\pi + \frac{\pi}{2} - 5\pi \sin 	heta$ for $n \in \mathbb{Z}$.Dividing by $\pi$,we get $5 \cos 	heta + 5 \sin 	heta = n + \frac{1}{2}$,which simplifies to $\cos 	heta + \sin 	heta = \frac{2n + 1}{10}$.Multiplying by $\frac{1}{\sqrt{2}}$,we get $\sin(	heta + \frac{\pi}{4}) = \frac{2n + 1}{10\sqrt{2}}$.Since $-1 \le \sin(	heta + \frac{\pi}{4}) \le 1$,we have $-10\sqrt{2} \le 2n + 1 \le 10\sqrt{2}$.Given $\sqrt{2} \approx 1.414$,$10\sqrt{2} \approx 14.14$,so $-14.14 \le 2n + 1 \le 14.14$.$-15.14 \le 2n \le 13.14$,which implies $-7.57 \le n \le 6.57$.Thus,$n \in \{-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$,which gives $14$ possible values for $n$.For each value of $n$,the equation $\sin(	heta + \frac{\pi}{4}) = k$ (where $|k| Therefore,the total number of solutions is $14 	imes 2 = 28$.

The number of solutions of $\tan(5\pi \cos \theta) = \cot(5\pi \sin \theta)$ for $\theta$ in $(0, 2\pi)$ is:

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