If sin theta = sqrt{3} cos theta and -pi

Question

(B) Given $\sin 	heta = \sqrt{3} \cos 	heta$.Dividing both sides by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$),we get $	an 	heta = \sqrt{3}$.We

Vedclass · Accepted Answer

$-\frac{4\pi}{6}$

Vedclass · Answer

(B) Given $\sin 	heta = \sqrt{3} \cos 	heta$.Dividing both sides by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$),we get $	an 	heta = \sqrt{3}$.We know that $	an \frac{\pi}{3} = \sqrt{3}$.The general solution for $	an 	heta = 	an \alpha$ is $	heta = n\pi + \alpha$,where $n \in \mathbb{Z}$.So,$	heta = n\pi + \frac{\pi}{3}$.We are given the interval $-\pi For $n = -1$,$	heta = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$.To match the given options,we can write $-\frac{2\pi}{3}$ as $-\frac{4\pi}{6}$.Thus,the correct option is $B$.

If $\sin \theta = \sqrt{3} \cos \theta$ and $-\pi < \theta < 0$,then $\theta = $

Similar Questions

The number of solutions of $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$ is equal to:

The number of values of $x$ for which $\sin 2x + \sin 4x = 2$ is

The general solution of $\sin x + \cos x = 1$ is

The number of solutions of $\sec x \cos 5x + 1 = 0$ in the interval $[0, 2\pi]$ is

If $\alpha, \beta, \gamma$ and $\delta$ are the solutions of the equation $\tan \left( \theta + \frac{\pi}{4} \right) = 3 \tan 3\theta$,no two of which have equal tangents,then the value of $\tan \alpha + \tan \beta + \tan \gamma + \tan \delta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module