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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Find the general solution of the equation $\sec^{2} 2x = 1 - \tan 2x$.

Number of solutions of the equation $\tan^2 x + 3 \cot^2 x = 2 \sec^2 x$ lying in the interval $[0, 2\pi]$ is

The number of solutions of the equation $4 \cos 2 \theta \cos 3 \theta = \sec \theta$ in the interval $[0, 2 \pi]$ is

Solve $\tan 2x = -\cot \left(x + \frac{\pi}{3}\right)$

If $\cot (\alpha + \beta ) = 0,$ then $\sin (\alpha + 2\beta ) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module