If sqrt{3} cos theta + sin theta 0,then: | vedclass

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Question

(B) Given the inequality $\sqrt{3} \cos 	heta + \sin 	heta > 0$. Divide the entire inequality by $2$: $\frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2}

Vedclass · Accepted Answer

$-\frac{\pi}{3} < 	heta < \frac{2 \pi}{3}$

Vedclass · Answer

(B) Given the inequality $\sqrt{3} \cos 	heta + \sin 	heta > 0$. Divide the entire inequality by $2$: $\frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta > 0$. We know that $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{6} = \frac{1}{2}$. So,$\cos \frac{\pi}{6} \cos 	heta + \sin \frac{\pi}{6} \sin 	heta > 0$. Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we get: $\cos(	heta - \frac{\pi}{6}) > 0$. For $\cos x > 0$,$x$ must lie in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ within one period. So,$-\frac{\pi}{2} Adding $\frac{\pi}{6}$ to all parts: $-\frac{\pi}{2} + \frac{\pi}{6} $-\frac{3\pi}{6} + \frac{\pi}{6} $-\frac{2\pi}{6} $-\frac{\pi}{3} Thus,the correct option is $B$.

If $\sqrt{3} \cos \theta + \sin \theta > 0$,then:

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