If angle theta in [0, 2pi] satisfies both the | vedclass

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

Question

(B) Given equations are $\cot 	heta = \sqrt{3}$ and $\sqrt{3} \sec 	heta + 2 = 0$. From $\cot 	heta = \sqrt{3}$,we have $	an 	heta = \frac{1}{\sq

Vedclass · Accepted Answer

$\frac{7 \pi}{6}$

Vedclass · Answer

(B) Given equations are $\cot 	heta = \sqrt{3}$ and $\sqrt{3} \sec 	heta + 2 = 0$. From $\cot 	heta = \sqrt{3}$,we have $	an 	heta = \frac{1}{\sqrt{3}}$. This implies $	heta$ is in the $1^{	ext{st}}$ or $3^{	ext{rd}}$ quadrant. From $\sqrt{3} \sec 	heta + 2 = 0$,we have $\sec 	heta = -\frac{2}{\sqrt{3}}$,which means $\cos 	heta = -\frac{\sqrt{3}}{2}$. Since $\cos 	heta$ is negative,$	heta$ must be in the $2^{	ext{nd}}$ or $3^{	ext{rd}}$ quadrant. Combining both conditions,$	heta$ must lie in the $3^{	ext{rd}}$ quadrant. In the $3^{	ext{rd}}$ quadrant,$	an 	heta = \frac{1}{\sqrt{3}}$ corresponds to $	heta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

If angle $\theta$ in $[0, 2\pi]$ satisfies both the equations $\cot \theta = \sqrt{3}$ and $\sqrt{3} \sec \theta + 2 = 0$,then $\theta$ is equal to

Similar Questions

The number of solutions of the equation $\cos^2 2x + \cos^2 \frac{5x}{4} = \cos 2x \cos^2 5x$ in the interval $[0, \frac{\pi}{3}]$ is:

If $\frac{\tan 3\theta - 1}{\tan 3\theta + 1} = \sqrt{3}$,then the general value of $\theta$ is

The general solution of $\cos 4\theta = \cos 3\theta$ is $\theta =.....$

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

The number of solutions of the equation $2x + 3 \tan x = \pi$,where $x \in [-2\pi, 2\pi] - \left\{ \pm \frac{\pi}{2}, \pm \frac{3\pi}{2} \right\}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module