Question

If $ y\sin \theta = 3,{\text{ and}}$ $y = 4(1 + \sin \theta ),0 \leqslant \theta \leqslant 2\pi $ .Then, $\theta = ...........$
\[
  {\text{(A) }}\dfrac{\pi }{6},\dfrac{{5\pi }}{6} \\
  {\text{(B) }}\dfrac{\pi }{3},\dfrac{{2\pi }}{3} \\
  {\text{(C) }}\dfrac{\pi }{4},\dfrac{{5\pi }}{4} \\
  {\text{(D) }}\dfrac{\pi }{2},\pi \\
 \]

Answer 1

Hint:This is a problem related to Trigonometric equations. Two variables and two equations can be solved easily by replacing one variable in terms of another variable from one equation to the other. This will give a linear or quadratic equation with one variable.

Complete step-by-step answer:
Firstly, rewrite the equations given in the question as,
\[
  y\sin \theta = 3{\text{ }}.................{\text{ (1)}} \\
  y = 4(1 + \sin \theta ){\text{ }}.........{\text{ (2)}} \\
 \]
From equation (1), we can get the value of $\sin \theta $ as
$\sin \theta = \dfrac{3}{y}$
Putting this value of $\sin \theta $ in equation (2), we get
$ \Rightarrow y = 4 + 4 \times \dfrac{3}{y}$
After simplification, we will get a quadratic equation in terms of $y$ as
$
   \Rightarrow y = 4 + 4 \times \dfrac{3}{y} \\
   \Rightarrow {y^2} = 4y + 12 \\
   \Rightarrow {y^2} - 4y - 12 = 0 \\
 $
After factorising the above equation we get
$
   \Rightarrow {y^2} - 6y + 2y - 12 = 0 \\
   \Rightarrow y(y - 6) + 2(y - 6) = 0 \\
   \Rightarrow (y - 6)(y + 2) = 0 \\
 $
This will give two values of $y$ as
$ \Rightarrow y = 6, - 2$
Now putting these values of $y$ in equation (1), we will get,
$
  {\text{when }}y = 6, \\
  \sin \theta = \dfrac{3}{6} = \dfrac{1}{2} \\
 $
And putting the other value of $y$ in equation (2), we get
$
  {\text{when }}y = - 2, \\
  4(1 + \sin \theta ) = - 2 \\
   \Rightarrow (1 + \sin \theta ) = - \dfrac{2}{4} = - \dfrac{1}{2} \\
   \Rightarrow \sin \theta = - \dfrac{3}{2} \\
 $
As we know that $ - 1 \leqslant \sin \theta \leqslant 1$, so this value of $\sin \theta $ can be neglected. Therefore, we have one value of$\sin \theta $, which is equal to $\dfrac{1}{2}$.
Therefore, the value of $\theta $ will be equal to $\dfrac{\pi }{6}$ and value of it in the 2nd quadrant is $\pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}$.
This way, there are two values of $\theta $ are $\dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{6}$ which satisfies both the equations in the question.

So, the correct answer is “Option A”.

Note:Solving these types of problems, you should be well aware of the range of $\sin \theta $. It should be noted here that we have converted the equation into a quadratic equation with one variable $y$ and solved. It is noted here that in the given range of $\theta $, there are two instances where $\sin \theta $ has the same value, i.e., $\dfrac{1}{2}$.