Question

The number of real solutions of the trigonometric equation $\sin 3\theta =4\sin \theta \sin 2\theta \sin 4\theta $ in $0\le \theta \le \pi $ is.
(a) 2 real solutions
(b) 4 real solutions
(c) 6 real solutions
(d) 8 real solutions

Answer 1

Hint:For solving this question we will use some trigonometric formulae like $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ , $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and then find the value of $\theta $ which will satisfy the given equation. After finding the solutions we will just count them and find the correct answer.

Complete step-by-step answer:
Given:
It is given that $\sin 3\theta =4\sin \theta \sin 2\theta \sin 4\theta $ in $0\le \theta \le \pi $ and we have to find the number of real solutions.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & 2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)...................\left( 1 \right) \\
 & 2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)...................\left( 2 \right) \\
 & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)...............\left( 3 \right) \\
 & \cos \dfrac{2\pi }{3}=-\dfrac{1}{2}............................................................\left( 4 \right) \\
\end{align}$
We have,
$\sin 3\theta =4\sin \theta \sin 2\theta \sin 4\theta $
Now, using the formula from the equation (1) to write $2\sin 2\theta \sin \theta =\cos \theta -\cos 3\theta $ . Then,
$\begin{align}
  & \sin 3\theta =4\sin \theta \sin 2\theta \sin 4\theta \\
 & \Rightarrow \sin 3\theta =2\sin 4\theta \left( 2\sin 2\theta \sin \theta \right) \\
 & \Rightarrow \sin 3\theta =2\sin 4\theta \left( \cos \left( 2\theta -\theta \right)-\cos \left( 2\theta +\theta \right) \right) \\
 & \Rightarrow \sin 3\theta =2\sin 4\theta \left( \cos \theta -\cos 3\theta \right) \\
 & \Rightarrow \sin 3\theta =2\sin 4\theta \cos \theta -2\sin 4\theta \cos 3\theta \\
\end{align}$
Now, using the formula from the equation (2) to write $2\sin 4\theta \cos \theta =\sin 5\theta +\sin 3\theta $ and $2\sin 4\theta \cos 3\theta =\sin 7\theta +\sin \theta $ . Then,
$\begin{align}
  & \sin 3\theta =2\sin 4\theta \cos \theta -2\sin 4\theta \cos 3\theta \\
 & \Rightarrow \sin 3\theta =\sin \left( 4\theta +\theta \right)+\sin \left( 4\theta -\theta \right)-\sin \left( 4\theta +3\theta \right)-\sin \left( 4\theta -3\theta \right) \\
 & \Rightarrow \sin 3\theta =\sin 5\theta +\sin 3\theta -\sin 7\theta -\sin \theta \\
 & \Rightarrow \sin 7\theta -\sin 5\theta +\sin \theta =0 \\
\end{align}$
Now, using the formula from the equation (3) to write $\sin 7\theta -\sin 5\theta =2\cos 6\theta \sin \theta $ . Then,
$\begin{align}
  & \sin 7\theta -\sin 5\theta +\sin \theta =0 \\
 & \Rightarrow 2\cos \left( \dfrac{7\theta +5\theta }{2} \right)\sin \left( \dfrac{7\theta -5\theta }{2} \right)+\sin \theta =0 \\
 & \Rightarrow 2\cos 6\theta \sin \theta +\sin \theta =0 \\
 & \Rightarrow 2\sin \theta \left( \cos 6\theta +\dfrac{1}{2} \right)=0 \\
\end{align}$
Now, write $\dfrac{1}{2}=-\cos \dfrac{2\pi }{3}$ in the above equation. Then,
$\begin{align}
  & 2\sin \theta \left( \cos 6\theta +\dfrac{1}{2} \right)=0 \\
 & \Rightarrow 2\sin \theta \left( \cos 6\theta -\cos \dfrac{2\pi }{3} \right)=0 \\
 & \Rightarrow \sin \theta =0\text{ , }\cos 6\theta =\cos \dfrac{2\pi }{3} \\
\end{align}$
Now, from the above result we have the following two equations:
$\begin{align}
  & \sin \theta =0......................\left( 5 \right) \\
 & \cos 6\theta =\cos \dfrac{2\pi }{3}..........\left( 6 \right) \\
\end{align}$
Now, we will find suitable values of $\theta $ for each of the above equation separately.
Form equation (5) we know that, $\sin \theta =0$ . And before we proceed, we should know that if $\sin \theta =0$ , then $\theta =n\pi $ where $n$ is any integer. But before we find $\theta $ , we should remember that $\theta $ can take values from $0$ to $\pi $ including $0$ & $\pi $ . Then,
$\begin{align}
  & \sin \theta =0 \\
 & \Rightarrow \theta =0,\pi ......................\left( 7 \right) \\
\end{align}$
Now, from the above result, suitable values of $\theta $ will be $0,\pi $ .
Now, before we proceed we should know one important result which we will use here.
If $\cos x=\cos y$ , then the general solution for $x$ in terms of y can be written as,
$x=2n\pi \pm y............\left( 8 \right)$ , where $n$ is any integer.
From equation (6) we know that $\cos 6\theta =\cos \dfrac{2\pi }{3}$ . So, we can use the formula from the equation (8). Then,
$\begin{align}
  & \cos 6\theta =\cos \dfrac{2\pi }{3} \\
 & \Rightarrow 6\theta =2n\pi \pm \dfrac{2\pi }{3} \\
 & \Rightarrow \theta =\dfrac{n\pi }{3}\pm \dfrac{\pi }{9} \\
\end{align}$
First, consider, $x=\dfrac{n\pi }{3}+\dfrac{\pi }{9}$
Put, $n=0$ . Then,
$\begin{align}
  & x=0+\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{\pi }{9}...........\left( 9 \right) \\
\end{align}$
Put, $n=1$ . Then,
$\begin{align}
  & x=\dfrac{\pi }{3}+\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{4\pi }{9}..........\left( 10 \right) \\
\end{align}$
Put, $n=2$ . Then,
$\begin{align}
  & x=\dfrac{2\pi }{3}+\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{7\pi }{9}.............\left( 11 \right) \\
\end{align}$
Put, $n=3$ . Then,
$\begin{align}
  & x=\dfrac{3\pi }{3}+\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{10\pi }{9}\left( >\pi \right) \\
\end{align}$
Now consider, $x=\dfrac{n\pi }{3}-\dfrac{\pi }{9}$
Put, $n=1$ . Then,
$\begin{align}
  & x=\dfrac{\pi }{3}-\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{2\pi }{9}.............\left( 12 \right) \\
\end{align}$
Put, $n=2$ . Then,
$\begin{align}
  & x=\dfrac{2\pi }{3}-\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{5\pi }{9}............\left( 13 \right) \\
\end{align}$
Put, $n=3$ . Then,
$\begin{align}
  & x=\dfrac{3\pi }{3}-\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{8\pi }{9}............\left( 14 \right) \\
\end{align}$
Put, $n=4$ . Then,
$\begin{align}
  & x=\dfrac{4\pi }{3}-\dfrac{\pi }{9} \\
 & \Rightarrow x=\dfrac{11\pi }{9}\left( >\pi \right) \\
\end{align}$
In the above calculations, we neglected the values which are greater than $\pi $ or lesser than 0.
From (7), (9), (10), (11), (12), (13) and (14) we have:
$x=0,\pi \text{ ; }x=\dfrac{\pi }{9}\text{ ; }x=\dfrac{4\pi }{9}\text{ ; }x=\dfrac{7\pi }{9}\text{ ; }x=\dfrac{2\pi }{9}\text{ ; }x=\dfrac{5\pi }{9}\text{ ; }x=\dfrac{8\pi }{9}$
Now, from the above result, it is evident that there will be total 8 suitable values of $\theta $ Hence, (d) will be the correct option.

Note: Here, we should use each formula which is mentioned in the solution. And for finding the value of $\theta $ and we should remember that $\theta $ can take values from $0$ to $\pi $ including $0$ & $\pi $ while solving. Then, find the number of suitable values and match the correct option. Moreover, we should avoid calculation mistakes while solving to get the correct answer.