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Question: Find the values of \( \theta \) which satisfy \( r\sin \theta = 3 \) and \( r = 4\left( {1 + \sin \t...

Find the values of θ\theta which satisfy rsin⁡θ=3r\sin \theta = 3 and r=4(1+sin⁡θ)r = 4\left( {1 + \sin \theta } \right) , 0≤θ≤2π0 \le \theta \le 2\pi
This question has multiple correct options
(A) θ=π6\theta = \dfrac{\pi }{6}
(B) θ=5π6\theta = \dfrac{{5\pi }}{6}
(C) θ=π3\theta = \dfrac{\pi }{3}
(D) θ=2π3\theta = \dfrac{{2\pi }}{3}

Explanation

Solution

Hint : In this question we have two expressions given in terms of rr and θ\theta . We have to find the value of θ\theta by eliminating rr from the expression by using the formula and the condition. The value of θ\theta lies between an interval defined by values 00 and 2π2\pi .

Complete step-by-step answer :
Given:
The first equation given is –
rsin⁡θ=3r\sin \theta = 3
And, the second equation given is –
r=4(1+sin⁡θ)r = 4\left( {1 + \sin \theta } \right)
Also, the value of θ\theta lies in the interval 0≤θ≤2π0 \le \theta \le 2\pi
To eliminate rr from the equation we put the value of rr from the second equation to the first equation and solve, we get,
⇒4(1+sin⁡θ)×sin⁡θ=3 ⇒4sin⁡2θ+4sin⁡θ−3=0 \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 θ\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 4sin⁡θ4\sin \theta and its coefficient is also 4.
The last part is the constant value and its value is −3- 3 .
We can solve this question in the following steps –
⇒4sin⁡2θ+4sin⁡θ−3=0 ⇒4sin⁡2θ−2sin⁡θ+6sin⁡θ−3=0 ⇒2sin⁡θ(2sin⁡θ−1)+3(2sin⁡θ−1)=0 ⇒(2sin⁡θ−1)(2sin⁡θ+3)=0 \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 θ\theta are –
⇒(2sin⁡θ−1)=0 ⇒sin⁡θ=12 ⇒θ=π6or5π6 \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⁡θ\sin \theta lies in the first and third quadrant and satisfies the condition 0≤θ≤2π0 \le \theta \le 2\pi .
And similarly,
⇒(2sin⁡θ+3)=0 ⇒sin⁡θ=−32 \Rightarrow \left( {2\sin \theta + 3} \right) = 0\\\ \Rightarrow \sin \theta = \dfrac{{ - 3}}{2}
But it's not possible because the value of sin⁡θ\sin \theta cannot be less than zero. So, neglecting this value we get the value of θ\theta as –
⇒θ=π6or5π6\Rightarrow \theta = \dfrac{\pi }{6}{\rm{or }}\dfrac{{5\pi }}{6}

**Therefore, the correct answers are –
(A) θ=π6\theta = \dfrac{\pi }{6}
(B) θ=5π6\theta = \dfrac{{5\pi }}{6} **

Note : It should be noted that θ\theta has two values because according to the given condition the value of θ\theta must lie between the interval 0to2π0{\rm{ to 2}}\pi and the value of sin⁡θ\sin \theta has the same value in two quadrants, that is the reason the value of satisfies both the condition and has two values.