Question

If \[3\sin 2\theta =2\sin 3\theta \] where \[0<\theta <\pi \] , then the value of \[\sin \theta \] is
(a) \[\dfrac{\sqrt{2}}{3}\]
(b) \[\dfrac{\sqrt{3}}{\sqrt{5}}\]
(c) \[\dfrac{\sqrt{15}}{4}\]
(d) \[\dfrac{\sqrt{2}}{\sqrt{5}}\]

Answer 1

Hint: To solve this question we will use formulas of basic trigonometric identities, one of them is \[\sin \theta =\sqrt{(1-{{\cos }^{2}}\theta }\] where \[\theta \] is the given angle. Rearranging the terms given in the form \[3\sin 2\theta =2\sin 3\theta \] where \[0 < \theta < \pi \] we get the required result as the value of \[\sin \theta \] .

Complete step-by-step answer:
Given \[3\sin 2\theta =2\sin 3\theta \] where \[0<\theta <\pi \]
To calculate the value of \[\sin \theta \] .
We have a trigonometric formula given as \[\sin 2\theta =2\sin \theta \cos \theta \] , we will substitute this value in the left hand side of the equation \[3\sin 2\theta =2\sin 3\theta \] .
Similarly using the trigonometric formula given as \[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \] and substituting this in the right hand side of the given equation \[3\sin 2\theta =2\sin 3\theta \] , we will get our required angles.
Therefore, we have \[3\sin 2\theta =2\sin 3\theta \]

 \[\begin{align}
  & \Rightarrow 3\times 2sin\theta cos\theta =2(3sin\theta -4si{{n}^{3}}\theta ) \\
 & \Rightarrow 6sin\theta cos\theta =6sin\theta -8si{{n}^{3}}\theta \\
 & \Rightarrow sin\theta (6cos\theta -6+8si{{n}^{2}}\theta )=0 \\
 & \Rightarrow sin\theta =0~or~3cos\theta +4si{{n}^{2}}\theta =3 \\
 & \Rightarrow sin\theta =0~or~3cos\theta +4(1-co{{s}^{2}}\theta )=3 \\
\end{align}\]
 \[\begin{align}
  & \Rightarrow sin\theta =0~,3cos\theta -4co{{s}^{2}}\theta +1=0 \\
 & \Rightarrow 4co{{s}^{2}}\theta -3cos\theta -1=0 \\
 & \Rightarrow 4co{{s}^{2}}\theta -4cos\theta +cos\theta -1=0 \\
 & ~\Rightarrow 4cos\theta (cos\theta -1)+1(cos\theta -1)=0 \\
 & \Rightarrow (4cos\theta +1)(cos\theta -1)=0 \\
 & \Rightarrow cos\theta =\dfrac{-1}{4}~or~cos\theta =1 \\
\end{align}\]
Assuming the value of \[cos\theta =1\] and by applying the formula \[\sin \theta =\sqrt{(1-{{\cos }^{2}}\theta )}\] we get,
 \[\begin{align}
  & \sin \theta =0 \\
 & \Rightarrow \theta =0 \\
\end{align}\]
 \[\theta =0\] is not in our range as we had \[0<\theta <\pi \] .
Hence we need to take the other obtained value of cos.
Assuming the value of \[cos\theta =\dfrac{-1}{4}\] and by applying the formula \[\sin \theta =\sqrt{(1-{{\cos }^{2}}\theta )}\] we get,
 \[\begin{align}
  & \Rightarrow sin\theta =\sqrt{1-{{\left( \dfrac{-1}{4} \right)}^{2}}} \\
 & \Rightarrow sin\theta =\sqrt{1-\dfrac{1}{16}} \\
 & \Rightarrow sin\theta =\dfrac{\sqrt{15}}{4} \\
\end{align}\]
Hence, we obtain \[sin\theta =\dfrac{\sqrt{15}}{4}\] which is option (c).

So, the correct answer is “Option (c)”.

Note: A minute but very important error in the question could be assuming the value of \[cos\theta =1\]
rather than going for the value as \[cos\theta =\dfrac{-1}{4}\] which gives the result of \[\sin \theta =0\] \[\Rightarrow \theta =0\] , which is wrong because we had given value of \[\theta \] lies in between 0 and \[\pi \] or \[0<\theta <\pi \] and not exactly equal to 0. Hence it will lead to a wrong result. So, it is very important to check which value of sin or cos is defined or valid according to the given range of \[\theta \] in the question.