Question

If we have a trigonometric equation $\sin 3\theta =\sin \theta $ , how many solutions exist such that $-2\pi < \theta < 2\pi $ ?
A. 11
B. 8
C. 7
D. 5

Answer 1

Hint: To find the number of solutions for $\sin 3\theta =\sin \theta $ such that $-2\pi <\theta <2\pi $ , we will find its general solution. We can write the given equation as $\sin 3\theta -\sin \theta =0$ . Using $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ , we can write $2\cos \left( \dfrac{3\theta +\theta }{2} \right)\sin \left( \dfrac{3\theta -\theta }{2} \right)=0$ . Solving this results in $\cos 2\theta =0\text{ or }\sin \theta =0$ . The general solution will be $\theta =\left( 2n+1 \right)\dfrac{\pi }{4},n\pi $ . By substituting the values for $n=Z$ , where Z is an integer, and considering the resulting values that are in $-2\pi <\theta <2\pi $ , we will get the required number of solutions.

Complete step-by-step solution
It is given that $\sin 3\theta =\sin \theta $ . We have to find the number of solutions that exists in $-2\pi <\theta <2\pi $ . We will have to find its general solution.
We have, $\sin 3\theta =\sin \theta $
Let us take $\sin \theta $ to the LHS. We will get
$\sin 3\theta -\sin \theta =0$
We know that $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ . Hence, we can write the above equation as
$2\cos \left( \dfrac{3\theta +\theta }{2} \right)\sin \left( \dfrac{3\theta -\theta }{2} \right)=0$
Let’s solve the angles. We will get
$2\cos \left( \dfrac{4\theta }{2} \right)\sin \left( \dfrac{2\theta }{2} \right)=0$
$\Rightarrow 2\cos \left( 2\theta \right)\sin \left( \theta \right)=0$
We can solve the above equation by taking 2 to the RHS. That is,
$\cos \left( 2\theta \right)\sin \left( \theta \right)=0$
From the above equation, we can understand that $\cos \left( 2\theta \right)\sin \left( \theta \right)=0$ if either of the following are 0:
$\cos 2\theta =0\text{ or }\sin \theta =0$
Let us consider $\cos 2\theta =0\text{ }$.
We know that if $\cos \theta =0$ then $\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$ .
Here, we have $\theta =2\theta $ . Hence, we can write the general solution as
$2\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$
Taking 2 from LHS to RHS, we will get
$\theta =\left( 2n+1 \right)\dfrac{\pi }{4}$
This means that \[\theta =...,\dfrac{-7\pi }{4},\dfrac{-5\pi }{4},\dfrac{-3\pi }{4},\dfrac{-\pi }{4},\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4},..\] for $n=Z$ , where Z is an integer.
Now, let us consider $\sin \theta =0$
We know that the general solution of $\sin \theta =0$ is $n\pi $ . Hence,
$\begin{align}
  & \sin \theta =0 \\
 & \Rightarrow \theta =n\pi \\
\end{align}$
We can write this as $\theta =...,-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi ,...$ for $n=Z$ .
It is given that $-2\pi <\theta <2\pi $ . So we will find the values of $\theta $ between $-2\pi \text{ and }2\pi $ , that is between $-{{360}^{\circ }}\text{ and }{{360}^{\circ }}$ .
From $\theta =\left( 2n+1 \right)\dfrac{\pi }{4}$ , we have \[\theta =...,\dfrac{-7\pi }{4},\dfrac{-5\pi }{4},\dfrac{-3\pi }{4},\dfrac{-\pi }{4},\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4},..\] . Let us convert these into degrees so that you will understand clearly which values are between $-2\pi \text{ and }2\pi $ .
We know that $1\text{ rad}=\dfrac{180}{\pi }\text{ degree}$ .
Let us now do this for each value.
$\dfrac{\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{\pi }{4}={{45}^{\circ }}$
$\dfrac{3\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{3\pi }{4}={{135}^{\circ }}$
$\dfrac{5\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{5\pi }{4}={{225}^{\circ }}$
$\dfrac{7\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{7\pi }{4}={{315}^{\circ }}$
$\dfrac{9\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{9\pi }{4}={{405}^{\circ }}$
We can see that $\dfrac{9\pi }{4}$ is greater than $2\pi $ . Hence, the angles from $\dfrac{9\pi }{4}$ and above will be greater than $2\pi $ . This will be also the case with the negative values. So we will consider only the values \[\theta =\dfrac{-7\pi }{4},\dfrac{-5\pi }{4},\dfrac{-3\pi }{4},\dfrac{-\pi }{4},\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4},\dfrac{7\pi }{4}\] .
Now let us take $\theta =...,-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi ,...$ . We know that within $-2\pi \text{ and }2\pi $ , the values will be
$\theta =-\pi ,0,\pi $
Thus, altogether, we will have the values \[\theta = -\pi ,\dfrac{-7\pi }{4},\dfrac{-5\pi }{4},\dfrac{-3\pi }{4},\dfrac{-\pi }{4},0,\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4},\dfrac{7\pi }{4},\pi \]
Hence, if $\sin 3\theta =\sin \theta $ , there exists 11 solutions such that $-2\pi <\theta <2\pi $. Hence, the correct option is A.

Note: When we found the values within the interval $-2\pi <\theta <2\pi $ , we did not consider $-2\pi \text{ and }2\pi $ . This is because the inequality sign. If the interval was $-2\pi \le \theta \le 2\pi $ , we would have considered $-2\pi \text{ and }2\pi $ . Do not write the formula of radian as $1\text{ rad}=\dfrac{\pi }{180}\text{ degree}$ . This would result in wrong degrees and hence wrong solutions.