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

If $\alpha$ is a root of the equation $25 \cos^2 \theta + 5 \cos \theta - 12 = 0$,for $\frac{\pi}{2} < \alpha < \pi$,then $\sin 2\alpha =$

The number of principal solutions of the equation $\tan 3x - \tan 2x - \tan x = 0$ is:

If the general solution of $\sin 5x = \cos 2x$ is of the form $x = a_n \cdot \frac{\pi}{2}$ for $n = 0, \pm 1, \pm 2, \dots$,then $a_n =$

If $x \neq (2n+1) \frac{\pi}{4}$,then the general solution of $\cos x + \cos 3x = \sin x + \sin 3x$ is

The general solution of $\sin^2 \theta \sec \theta + \sqrt{3} \tan \theta = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module