The equation r^2 cos^2 left(theta - frac{pi}{ | vedclass

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(D) Given equation: $r^2 \cos^2 \left(	heta - \frac{\pi}{3}ight) = 2$ Using the expansion $\cos(A - B) = \cos A \cos B + \sin A \sin B$: $r^2 \left

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a pair of straight lines

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(D) Given equation: $r^2 \cos^2 \left(	heta - \frac{\pi}{3}ight) = 2$ Using the expansion $\cos(A - B) = \cos A \cos B + \sin A \sin B$: $r^2 \left( \cos 	heta \cos \frac{\pi}{3} + \sin 	heta \sin \frac{\pi}{3} ight)^2 = 2$ $r^2 \left( \frac{1}{2} \cos 	heta + \frac{\sqrt{3}}{2} \sin 	heta ight)^2 = 2$ $r^2 \frac{1}{4} (\cos 	heta + \sqrt{3} \sin 	heta)^2 = 2$ $(r \cos 	heta + \sqrt{3} r \sin 	heta)^2 = 8$ Since $x = r \cos 	heta$ and $y = r \sin 	heta$: $(x + \sqrt{3} y)^2 = 8$ $(x + \sqrt{3} y)^2 - (2\sqrt{2})^2 = 0$ Using $a^2 - b^2 = (a - b)(a + b)$: $(x + \sqrt{3} y - 2\sqrt{2})(x + \sqrt{3} y + 2\sqrt{2}) = 0$ This represents a pair of straight lines.

The equation $r^2 \cos^2 \left(\theta - \frac{\pi}{3}\right) = 2$ represents

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