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Question: Number of solutions of equation \(\sin 9\theta =\sin \theta \) in the interval \(\left[ 0,2\pi \righ...

Number of solutions of equation sin⁡9θ=sin⁡θ\sin 9\theta =\sin \theta in the interval [0,2π]\left[ 0,2\pi \right] is?

Explanation

Solution

In the given question, we are given an equation in which we need to find the number of solutions. So, as we need to find the number of solutions, it is clear that the given equation has more than one solution. We will make use of some trigonometric identities in order to solve it.

Complete step by step answer:
According to the question, we are given an equation sin⁡9θ=sin⁡θ\sin 9\theta =\sin \theta and also given that θ\theta lies in the interval [0,2π]\left[ 0,2\pi \right]. Therefore, it is clear that the solution would lie in the interval [0,2π]\left[ 0,2\pi \right].
Now, let us consider the equation sin⁡9θ=sin⁡θ\sin 9\theta =\sin \theta as below:
sin⁡9θ=sin⁡θ ⇒sin⁡9θ−sin⁡θ=0 \begin{aligned} & \sin 9\theta =\sin \theta \\\ & \Rightarrow \sin 9\theta -\sin \theta =0 \\\ \end{aligned}
Now, making use of the trigonometric identity which is sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)
Now, our expression on left-hand side would be
sin⁡9θ−sin⁡θ=2cos⁡(9θ+θ2)sin⁡(9θ−θ2) ⇒2cos⁡5θsin⁡4θ \begin{aligned} & \sin 9\theta -\sin \theta =2\cos \left( \dfrac{9\theta +\theta }{2} \right)\sin \left( \dfrac{9\theta -\theta }{2} \right) \\\ & \Rightarrow 2\cos 5\theta \sin 4\theta \\\ \end{aligned}
Now, we need to equate the left-hand side of the expression to 0.
Now, here cos⁡5θ=0\cos 5\theta =0 and sin⁡4θ=0\sin 4\theta =0 .
Now,
⇒cos⁡5θ=0 ∴5θ=(2n+1)π2 ⇒θ=(2n+1)π10 \begin{aligned} & \Rightarrow \cos 5\theta =0 \\\ & \therefore 5\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{10} \\\ \end{aligned}
Now, taking n=0,1,2,3, … and now we need to substitute the values of n such that it doesn’t cross the interval.
Therefore, the values of θ\theta are:
⇒π10,3π10,5π10,7π10,9π10,11π10,13π10,15π10,17π10,19π10\Rightarrow \dfrac{\pi }{10},\dfrac{3\pi }{10},\dfrac{5\pi }{10},\dfrac{7\pi }{10},\dfrac{9\pi }{10},\dfrac{11\pi }{10},\dfrac{13\pi }{10},\dfrac{15\pi }{10},\dfrac{17\pi }{10},\dfrac{19\pi }{10}
So, here we get 10 values of theta.
Now,
⇒sin⁡4θ=0 ∴4θ=nπ ⇒θ=nπ4 \begin{aligned} & \Rightarrow \sin 4\theta =0 \\\ & \therefore 4\theta =n\pi \\\ & \Rightarrow \theta =\dfrac{n\pi }{4} \\\ \end{aligned}
Now, we can take all values of theta such that it does not cross the interval.
Therefore, the values of θ\theta are:
⇒π4,2π4,3π4,4π4,5π4,6π4,7π4,8π4\Rightarrow \dfrac{\pi }{4},\dfrac{2\pi }{4},\dfrac{3\pi }{4},\dfrac{4\pi }{4},\dfrac{5\pi }{4},\dfrac{6\pi }{4},\dfrac{7\pi }{4},\dfrac{8\pi }{4}
Therefore, by this there are 8 values of theta.
Therefore, total solutions of the given equation are 18.

Note: Now, the most important thing that we need to keep in mind is that we take the appropriate values of n and remember to check that theta taken is from the given interval or not. Sometimes, we forget to check the interval and randomly try to check the solution for every value of n.