If r sin theta = 3 and r = 4(1 + sin theta) f | vedclass

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(B) Given equations are $r \sin 	heta = 3$ and $r = 4(1 + \sin 	heta)$.Substitute $r = \frac{3}{\sin 	heta}$ into the second equation:$\frac{3}{\si

Vedclass · Accepted Answer

$\frac{\pi}{6}, \frac{5\pi}{6}$

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(B) Given equations are $r \sin 	heta = 3$ and $r = 4(1 + \sin 	heta)$.Substitute $r = \frac{3}{\sin 	heta}$ into the second equation:$\frac{3}{\sin 	heta} = 4(1 + \sin 	heta)$$3 = 4 \sin 	heta + 4 \sin^2 	heta$$4 \sin^2 	heta + 4 \sin 	heta - 3 = 0$Let $x = \sin 	heta$,then $4x^2 + 4x - 3 = 0$.$(2x - 1)(2x + 3) = 0$So,$x = \frac{1}{2}$ or $x = -\frac{3}{2}$.Since $-1 \le \sin 	heta \le 1$,we reject $x = -\frac{3}{2}$.Thus,$\sin 	heta = \frac{1}{2}$.For $0 \le 	heta \le 2\pi$,$	heta = \frac{\pi}{6}$ or $	heta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

If $r \sin \theta = 3$ and $r = 4(1 + \sin \theta)$ for $0 \le \theta \le 2\pi$,then $\theta = $

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