Question

Find the values of $ \theta $ which satisfy $ r\sin \theta = 3 $ and $ r = 4\left( {1 + \sin \theta } \right) $ , $ 0 \le \theta \le 2\pi $
This question has multiple correct options
(A) $ \theta = \dfrac{\pi }{6} $
(B) $ \theta = \dfrac{{5\pi }}{6} $
(C) $ \theta = \dfrac{\pi }{3} $
(D) $ \theta = \dfrac{{2\pi }}{3} $

Answer 1

Hint: In this question we have two expressions given in terms of $ r $ and $ \theta $ . We have to find the value of $ \theta $ by eliminating $ r $ from the expression by using the formula and the condition. The value of $ \theta $ lies between an interval defined by values $ 0 $ and $ 2\pi $ .

Complete step-by-step answer:
Given:
The first equation given is –
 $ r\sin \theta = 3 $
And, the second equation given is –
 $ r = 4\left( {1 + \sin \theta } \right) $
Also, the value of $ \theta $ lies in the interval $ 0 \le \theta \le 2\pi $
To eliminate $ r $ from the equation we put the value of $ r $ from the second equation to the first equation and solve, we get,
 $
\Rightarrow 4\left( {1 + \sin \theta } \right) \times \sin \theta = 3\\
\Rightarrow 4{\sin ^2}\theta + 4\sin \theta - 3 = 0
 $
This is a quadratic equation in terms of $ \theta $ . We can solve this step by step by the Grouping Method.
The first part of the equation is and its coefficient is 4.
The middle part of the equation is $ 4\sin \theta $ and its coefficient is also 4.
The last part is the constant value and its value is $ - 3 $ .
We can solve this question in the following steps –
 $
\Rightarrow 4{\sin ^2}\theta + 4\sin \theta - 3 = 0\\
\Rightarrow 4{\sin ^2}\theta - 2\sin \theta + 6\sin \theta - 3 = 0\\
\Rightarrow 2\sin \theta \left( {2\sin \theta - 1} \right) + 3\left( {2\sin \theta - 1} \right) = 0\\
\Rightarrow \left( {2\sin \theta - 1} \right)\left( {2\sin \theta + 3} \right) = 0
 $
So, the value of $ \theta $ are –
 $
\Rightarrow \left( {2\sin \theta - 1} \right) = 0\\
\Rightarrow \sin \theta = \dfrac{1}{2}\\
\Rightarrow \theta = \dfrac{\pi }{6}{\rm{ or }}\dfrac{{5\pi }}{6}
 $
Because this value of $ \sin \theta $ lies in the first and third quadrant and satisfies the condition $ 0 \le \theta \le 2\pi $ .
And similarly,
 $
\Rightarrow \left( {2\sin \theta + 3} \right) = 0\\
\Rightarrow \sin \theta = \dfrac{{ - 3}}{2}
 $
But it's not possible because the value of $ \sin \theta $ cannot be less than zero. So, neglecting this value we get the value of $ \theta $ as –
 $ \Rightarrow \theta = \dfrac{\pi }{6}{\rm{or }}\dfrac{{5\pi }}{6} $

Therefore, the correct answers are –
(A) $ \theta = \dfrac{\pi }{6} $
(B) $ \theta = \dfrac{{5\pi }}{6} $

Note: It should be noted that $ \theta $ has two values because according to the given condition the value of $ \theta $ must lie between the interval $ 0{\rm{ to 2}}\pi $ and the value of $ \sin \theta $ has the same value in two quadrants, that is the reason the value of satisfies both the condition and has two values.