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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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}$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

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

Similar Questions

${\left( \frac{\cos A + \cos B}{\sin A - \sin B} \right)^n} + {\left( \frac{\sin A + \sin B}{\cos A - \cos B} \right)^n}$ (where $n$ is an integer) $=$

If $\cos x + \cos y = -\cos \alpha$ and $\sin x + \sin y = -\sin \alpha$,then $\cot \left(\frac{x+y}{2}\right) = $

If $0 < x, y < \pi$ and $\cos x + \cos y - \cos(x + y) = \frac{3}{2}$,then $\sin x + \cos y$ is equal to ...... .

If $|\sin x-\cos ^2 x| \geq|3-3 \sin x+\sin ^2 x|+4|\sin x-1|$,then $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module