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Question: If \( y\sin \theta = 3,{\text{ and}}\) \(y = 4(1 + \sin \theta ),0 \leqslant \theta \leqslant 2\pi \...

If ysin⁡θ=3, and y\sin \theta = 3,{\text{ and}} y=4(1+sin⁡θ),0⩽θ⩽2πy = 4(1 + \sin \theta ),0 \leqslant \theta \leqslant 2\pi .Then, θ=...........\theta = ...........

(A) π6,5π6 (B) π3,2π3 (C) π4,5π4 (D) π2,π  {\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 \\\
Explanation

Solution

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,

ysin⁡θ=3 ................. (1) y=4(1+sin⁡θ) ......... (2)  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⁡θ\sin \theta as
sin⁡θ=3y\sin \theta = \dfrac{3}{y}
Putting this value of sin⁡θ\sin \theta in equation (2), we get
⇒y=4+4×3y\Rightarrow y = 4 + 4 \times \dfrac{3}{y}
After simplification, we will get a quadratic equation in terms of yy as
⇒y=4+4×3y ⇒y2=4y+12 ⇒y2−4y−12=0  \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
⇒y2−6y+2y−12=0 ⇒y(y−6)+2(y−6)=0 ⇒(y−6)(y+2)=0  \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 yy as
⇒y=6,−2\Rightarrow y = 6, - 2
Now putting these values of yy in equation (1), we will get,
when y=6, sin⁡θ=36=12  {\text{when }}y = 6, \\\ \sin \theta = \dfrac{3}{6} = \dfrac{1}{2} \\\
And putting the other value of yy in equation (2), we get
when y=−2, 4(1+sin⁡θ)=−2 ⇒(1+sin⁡θ)=−24=−12 ⇒sin⁡θ=−32  {\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⩽sin⁡θ⩽1 - 1 \leqslant \sin \theta \leqslant 1, so this value of sin⁡θ\sin \theta can be neglected. Therefore, we have one value ofsin⁡θ\sin \theta , which is equal to 12\dfrac{1}{2}.
Therefore, the value of θ\theta will be equal to π6\dfrac{\pi }{6} and value of it in the 2nd quadrant is π−π6=5π6\pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}.
This way, there are two values of θ\theta are π6\dfrac{\pi }{6} and 5π6\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⁡θ\sin \theta . It should be noted here that we have converted the equation into a quadratic equation with one variable yy and solved. It is noted here that in the given range of θ\theta , there are two instances where sin⁡θ\sin \theta has the same value, i.e., 12\dfrac{1}{2}.