Question

Number of solutions of the equation $\sin 7\theta =\sin \theta +\sin 3\theta $ for $\theta \in \left( 0,\pi \right)$ is equal to:
(a) 7
(b) 6
(c) 4
(d) 3

Answer 1

Hint: First of all, we are going to subtract $\sin \theta $ on both sides then we will apply the trigonometric identity on the left hand side of the equation. And the trigonometric identity is:$\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ then we will solve the given trigonometric equation and hence, will find the solutions of the equation.

Complete step-by-step answer:
The equation given in the above problem in which we have to find the number of solutions is:
$\sin 7\theta =\sin \theta +\sin 3\theta $
Subtracting $\sin \theta $ on both the sides we get,
$\sin 7\theta -\sin \theta =\sin 3\theta $
Now, as you can see that the left hand side of the above equation is written in the form of $\sin C-\sin D$ and we know that:
$\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$
Using the above trigonometric identity, we are going to solve the given trigonometric equation as follows:
$\begin{align}
  & 2\cos \left( \dfrac{7\theta +\theta }{2} \right)\sin \left( \dfrac{7\theta -\theta }{2} \right)=\sin 3\theta \\
 & \Rightarrow 2\cos 4\theta \sin 3\theta =\sin 3\theta \\
 & \Rightarrow 2\cos 4\theta \sin 3\theta -\sin 3\theta =0 \\
\end{align}$
Taking $\sin 3\theta $ as common in the above equation we get,
$\sin 3\theta \left( 2\cos 4\theta -1 \right)=0$
Now, equating $\sin 3\theta $ and $2\cos 4\theta -1$ to 0 separately we get,
$\begin{align}
  & \sin 3\theta =0.........Eq.(1) \\
 & 2\cos 4\theta -1=0 \\
 & \Rightarrow \cos 4\theta =\dfrac{1}{2}.......Eq.(2) \\
\end{align}$
We know that, $\sin \theta =0$ when $\theta =\pm n\pi $ and $\cos \theta =\dfrac{1}{2}$ when $\theta =2n\pi \pm \dfrac{\pi }{3}$ so substituting these values in the above eq. (1 & 2) we get,
$3\theta =\pm n\pi $
In the above equation, n can take values 0, 1, 2, 3…..
$3\theta =0,\pm \pi ,\pm 2\pi ,\pm 3\pi .....$
Dividing 3 on both the sides we get,
$\theta =0,\pm \dfrac{\pi }{3},\pm \dfrac{2\pi }{3},\pm \dfrac{3\pi }{3}.....$
Now, we are going to write the relationship between the angle of cosine and its value.
$4\theta =2n\pi \pm \dfrac{\pi }{3}$
Substituting the value of n as 0, 1, 2, 3…… we get,
$4\theta =0\pm \dfrac{\pi }{3},2\pi \pm \dfrac{\pi }{3},4\pi \pm \dfrac{\pi }{3}.....$
Writing solutions which have positive sign we get,
$\begin{align}
  & 4\theta =0+\dfrac{\pi }{3},2\pi +\dfrac{\pi }{3},4\pi +\dfrac{\pi }{3}....... \\
 & \Rightarrow 4\theta =\dfrac{\pi }{3},\dfrac{7\pi }{3},\dfrac{13\pi }{3},\dfrac{19\pi }{3}........ \\
 & \Rightarrow \theta =\dfrac{\pi }{12},\dfrac{7\pi }{12},\dfrac{13\pi }{12},\dfrac{19\pi }{12}......... \\
\end{align}$
Writing solutions which have negative sign we get,
$\begin{align}
  & 4\theta =0-\dfrac{\pi }{3},2\pi -\dfrac{\pi }{3},4\pi -\dfrac{\pi }{3}...... \\
 & \Rightarrow 4\theta =-\dfrac{\pi }{3},\dfrac{5\pi }{3},\dfrac{11\pi }{3},\dfrac{17\pi }{3}........... \\
 & \Rightarrow \theta =-\dfrac{\pi }{12},\dfrac{5\pi }{12},\dfrac{11\pi }{12},\dfrac{17\pi }{12}......... \\
\end{align}$
It is given that, we are asked to find the solutions of the given equation in the interval $\theta \in \left( 0,\pi \right)$ then the total solutions that we are getting are:
$\theta =\dfrac{\pi }{3},\dfrac{2\pi }{3},\dfrac{\pi }{12},\dfrac{7\pi }{12},\dfrac{5\pi }{12},\dfrac{11\pi }{12}$
From the above, we can see that 6 solutions are obtained.

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

Note: The mistake that could happen in the above problem is in writing the wrong general solutions for sine and cosine. Like, we have shown above the general solutions for sine and cosine as follows:
$\begin{align}
  & \sin \theta =0 \\
 & \Rightarrow \theta =\pm n\pi \\
\end{align}$
$\begin{align}
  & \cos \theta =\dfrac{\pi }{3} \\
 & \Rightarrow \theta =2n\pi \pm \dfrac{\pi }{3} \\
\end{align}$
The mistake that could happen is that you might forget to put a negative sign in the general formula. Doing this mistake will cost you the loss of a number of solutions.