Solveeit Logo
Login

Question

Question: If we have a trigonometric equation \(\sin 3\theta =\sin \theta \) , how many solutions exist such t...

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

Explanation

Solution

To find the number of solutions for sin⁡3θ=sin⁡θ\sin 3\theta =\sin \theta such that −2π<θ<2π-2\pi <\theta <2\pi , we will find its general solution. We can write the given equation as sin⁡3θ−sin⁡θ=0\sin 3\theta -\sin \theta =0 . Using sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) , we can write 2cos⁡(3θ+θ2)sin⁡(3θ−θ2)=02\cos \left( \dfrac{3\theta +\theta }{2} \right)\sin \left( \dfrac{3\theta -\theta }{2} \right)=0 . Solving this results in cos⁡2θ=0 or sin⁡θ=0\cos 2\theta =0\text{ or }\sin \theta =0 . The general solution will be θ=(2n+1)π4,nπ\theta =\left( 2n+1 \right)\dfrac{\pi }{4},n\pi . By substituting the values for n=Zn=Z , where Z is an integer, and considering the resulting values that are in −2π<θ<2π-2\pi <\theta <2\pi , we will get the required number of solutions.

Complete step-by-step solution
It is given that sin⁡3θ=sin⁡θ\sin 3\theta =\sin \theta . We have to find the number of solutions that exists in −2π<θ<2π-2\pi <\theta <2\pi . We will have to find its general solution.
We have, sin⁡3θ=sin⁡θ\sin 3\theta =\sin \theta
Let us take sin⁡θ\sin \theta to the LHS. We will get
sin⁡3θ−sin⁡θ=0\sin 3\theta -\sin \theta =0
We know that sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\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
2cos⁡(3θ+θ2)sin⁡(3θ−θ2)=02\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
2cos⁡(4θ2)sin⁡(2θ2)=02\cos \left( \dfrac{4\theta }{2} \right)\sin \left( \dfrac{2\theta }{2} \right)=0
⇒2cos⁡(2θ)sin⁡(θ)=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⁡(2θ)sin⁡(θ)=0\cos \left( 2\theta \right)\sin \left( \theta \right)=0
From the above equation, we can understand that cos⁡(2θ)sin⁡(θ)=0\cos \left( 2\theta \right)\sin \left( \theta \right)=0 if either of the following are 0:
cos⁡2θ=0 or sin⁡θ=0\cos 2\theta =0\text{ or }\sin \theta =0
Let us consider cos⁡2θ=0 \cos 2\theta =0\text{ }.
We know that if cos⁡θ=0\cos \theta =0 then θ=(2n+1)π2\theta =\left( 2n+1 \right)\dfrac{\pi }{2} .
Here, we have θ=2θ\theta =2\theta . Hence, we can write the general solution as
2θ=(2n+1)π22\theta =\left( 2n+1 \right)\dfrac{\pi }{2}
Taking 2 from LHS to RHS, we will get
θ=(2n+1)π4\theta =\left( 2n+1 \right)\dfrac{\pi }{4}
This means that θ=...,−7π4,−5π4,−3π4,−π4,π4,3π4,5π4,..\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=Zn=Z , where Z is an integer.
Now, let us consider sin⁡θ=0\sin \theta =0
We know that the general solution of sin⁡θ=0\sin \theta =0 is nπn\pi . Hence,
sin⁡θ=0 ⇒θ=nπ \begin{aligned} & \sin \theta =0 \\\ & \Rightarrow \theta =n\pi \\\ \end{aligned}
We can write this as θ=...,−3π,−2π,−π,0,π,2π,3π,...\theta =...,-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi ,... for n=Zn=Z .
It is given that −2π<θ<2π-2\pi <\theta <2\pi . So we will find the values of θ\theta between −2π and 2π-2\pi \text{ and }2\pi , that is between −360∘ and 360∘-{{360}^{\circ }}\text{ and }{{360}^{\circ }} .
From θ=(2n+1)π4\theta =\left( 2n+1 \right)\dfrac{\pi }{4} , we have θ=...,−7π4,−5π4,−3π4,−π4,π4,3π4,5π4,..\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 −2π and 2π-2\pi \text{ and }2\pi .
We know that 1 rad=180π degree1\text{ rad}=\dfrac{180}{\pi }\text{ degree} .
Let us now do this for each value.
π4 rad=180π ×π4=45∘\dfrac{\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{\pi }{4}={{45}^{\circ }}
3π4 rad=180π ×3π4=135∘\dfrac{3\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{3\pi }{4}={{135}^{\circ }}
5π4 rad=180π ×5π4=225∘\dfrac{5\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{5\pi }{4}={{225}^{\circ }}
7π4 rad=180π ×7π4=315∘\dfrac{7\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{7\pi }{4}={{315}^{\circ }}
9π4 rad=180π ×9π4=405∘\dfrac{9\pi }{4}\text{ rad}=\dfrac{180}{\pi }\text{ }\times \dfrac{9\pi }{4}={{405}^{\circ }}
We can see that 9π4\dfrac{9\pi }{4} is greater than 2π2\pi . Hence, the angles from 9π4\dfrac{9\pi }{4} and above will be greater than 2π2\pi . This will be also the case with the negative values. So we will consider only the values θ=−7π4,−5π4,−3π4,−π4,π4,3π4,5π4,7π4\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 θ=...,−3π,−2π,−π,0,π,2π,3π,...\theta =...,-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi ,... . We know that within −2π and 2π-2\pi \text{ and }2\pi , the values will be
θ=−π,0,π\theta =-\pi ,0,\pi
Thus, altogether, we will have the values θ=−π,−7π4,−5π4,−3π4,−π4,0,π4,3π4,5π4,7π4,π\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⁡3θ=sin⁡θ\sin 3\theta =\sin \theta , there exists 11 solutions such that −2π<θ<2π-2\pi <\theta <2\pi . Hence, the correct option is A.

Note: When we found the values within the interval −2π<θ<2π-2\pi <\theta <2\pi , we did not consider −2π and 2π-2\pi \text{ and }2\pi . This is because the inequality sign. If the interval was −2π≤θ≤2π-2\pi \le \theta \le 2\pi , we would have considered −2π and 2π-2\pi \text{ and }2\pi . Do not write the formula of radian as 1 rad=π180 degree1\text{ rad}=\dfrac{\pi }{180}\text{ degree} . This would result in wrong degrees and hence wrong solutions.