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 = $

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