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Question: Find the value of \(\theta \) which satisfy \(r\sin \theta =3\text{ and }r=4(1+\sin \theta ),0\le \t...

Find the value of θ\theta which satisfy rsin⁡θ=3 and r=4(1+sin⁡θ),0≤θ≤2πr\sin \theta =3\text{ and }r=4(1+\sin \theta ),0\le \theta \le 2\pi

Explanation

Solution

Initially we have to find the value of r using an algebraic method and then put that value in the given 2 equation. After that we have to use the given domain for the angle and analyze the possible values. By putting this value we can get the final answer.

Complete step-by-step answer:
We have to find the value of rr or just eliminate rr from the equation.
To do this we can use the elimination method as follows.
Given,
rsin⁡θ=3...(i) r=4(1+sinθ)........(ii) \begin{aligned} & r\sin \theta =3...(i) \\\ & r=4(1+sin\theta )........(ii) \\\ \end{aligned}
Putting the value of rr from (ii) to (i), we get:
4(1+sin⁡θ)sin⁡θ=3 4sin⁡θ+4sin⁡2θ−3=0 \begin{aligned} & 4(1+\sin \theta )\sin \theta =3 \\\ & 4\sin \theta +4{{\sin }^{2}}\theta -3=0 \\\ \end{aligned}
We can write above equation as :
4sin⁡2θ+6sin⁡θ−2sin⁡θ−3=04{{\sin }^{2}}\theta +6\sin \theta -2\sin \theta -3=0
Taking common terms out.
2sin⁡θ(2sin⁡θ+3)−1(2sin⁡θ+3)=02\sin \theta (2\sin \theta +3)-1(2\sin \theta +3)=0
Again, taking common terms out.
(2sin⁡θ−1)(2sin⁡θ+3)=0(2\sin \theta -1)(2\sin \theta +3)=0
Now we can solve each term independently and can get the value of θ\theta
2sin⁡θ−1=0 sin⁡θ=12 θ=sin⁡−1(12)  \begin{aligned} & 2\sin \theta -1=0 \\\ & \sin \theta =\dfrac{1}{2} \\\ & \theta ={{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\\ & \\\ \end{aligned}
Value of sin⁡30∘=12\sin {{30}^{\circ }}=\dfrac{1}{2}
Hence, one value of θ=30∘\theta ={{30}^{\circ }}
Similarly, we can solve for the other term.
2sin⁡θ+3=0 sin⁡θ=−32 \begin{aligned} & 2\sin \theta +3=0 \\\ & \sin \theta =-\dfrac{3}{2} \\\ \end{aligned}
This value of sin⁡\sin is not possible.

Therefore we get the value of θ=π6,5π6...\theta =\dfrac{\pi }{6},\dfrac{5\pi }{6}... so on.

Note: First thing to be noted in these types of questions is that we should avoid any extra term as much as possible. As in this question we have removed rr from the equation and then just solve the equation using the factor method. Another thing to be noted here is that we can also solve this question using the formula for finding the roots of a quadratic equation in which we have to calculate more and then solve to find the root. In these types of questions we should note that there may be infinite value as trigonometric functions are periodic.