The most general value of theta which satisfi | vedclass

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

Question

(B) Given trigonometric equations are $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Since $	an 	heta$ is negative and $\cos 	heta$ is p

Vedclass · Accepted Answer

$2 n \pi + \frac{7 \pi}{4}$

Vedclass · Answer

(B) Given trigonometric equations are $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Since $	an 	heta$ is negative and $\cos 	heta$ is positive,$	heta$ must lie in the fourth quadrant.The general solution for $	an 	heta = -1$ is $	heta = n \pi + \frac{3 \pi}{4}$.The general solution for $\cos 	heta = \frac{1}{\sqrt{2}}$ is $	heta = 2 n \pi \pm \frac{\pi}{4}$.For $	heta$ to satisfy both,we look for the angle in the fourth quadrant where $	an 	heta = -1$ and $\cos 	heta = \frac{1}{\sqrt{2}}$,which is $	heta = \frac{7 \pi}{4}$.Thus,the general solution is $	heta = 2 n \pi + \frac{7 \pi}{4}$.

The most general value of $\theta$ which satisfies both the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

Similar Questions

If $2 \cos^{2} \theta + 3 \cos \theta = 2$,then the permissible value of $\cos \theta$ is

If $\sin \theta + \cos \theta = 0$ and $0 < \theta < \pi$,then $\theta$ is:

If $2{\sin ^2}x + {\sin ^2}2x = 2$ for $-\pi < x < \pi$,then $x = $

The set $\{x \in R: \cos 2x + 2 \cos^2 x = 2\}$ is equal to

The number of distinct solutions of $\sec \theta + \tan \theta = \sqrt{3}$ for $0 \leqslant \theta \leqslant 2\pi$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module