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

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 $3\sin 2\theta =2\sin 3\theta$

& \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{aligned}$$ $$\begin{aligned} & \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{aligned}$$ Assuming the value of $$cos\theta =1$$ and by applying the formula $$\sin \theta =\sqrt{(1-{{\cos }^{2}}\theta )}$$ we get, $$\begin{aligned} & \sin \theta =0 \\\ & \Rightarrow \theta =0 \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ Hence, we obtain $$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.