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

Number of values of $x \in [0, 2\pi]$ satisfying the equation $\cot x - \cos x = 1 - \cot x \cos x$.

If $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$,then $\sin \left( \theta + \frac{\pi }{4} \right)$ equals

Find the general solution of $4 \cos 2x - 4 \sqrt{3} \sin 2x + \cos 3x - \sqrt{3} \sin 3x + \cos x - \sqrt{3} \sin x = 0$.

If $\tan (A - B) = 1$ and $\sec (A + B) = \frac{2}{\sqrt{3}}$,then the smallest positive value of $B$ is

The principal solutions of $(5+3 \sin \theta)(2 \cos \theta+1)=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module