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# Number of solutions of equation of $\sin 9\theta = \sin \theta$ in the interval $\left[ {0,2\pi } \right]$ isA. 16B. 17C. 18D. 15

Last updated date: 13th Jun 2024
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Hint: Given an equation, we have to find the number of solutions of the equation. Which means that there are several solutions of the equation and hence we have to find the general solution of the equation. The formula used here is the trigonometric sum to product formula, which is given by:
$\Rightarrow \sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$
If $\sin x = 0$, then $x$ would be a multiple of $\pi$.
If $\cos x = 0$, then $x$ would be an odd multiple of $\dfrac{\pi }{2}$.

Given the equation $\sin 9\theta = \sin \theta$, we have to find the number of solutions of $\theta$.
Also given that the $\theta$ should be within the interval $\left[ {0,2\pi } \right]$.
So the solutions of $\theta$ from the equation $\sin 9\theta = \sin \theta$ should be in interval $\left[ {0,2\pi } \right]$.
Consider the equation $\sin 9\theta = \sin \theta$, as given below:
$\Rightarrow \sin 9\theta = \sin \theta$
$\Rightarrow \sin 9\theta - \sin \theta = 0$
Apply the trigonometric sum to product formula to the L.H.S of the above equation, as given below:
$\Rightarrow \sin 9\theta - \sin \theta = 2\cos \left( {\dfrac{{9\theta + \theta }}{2}} \right)\sin \left( {\dfrac{{9\theta - \theta }}{2}} \right)$
$\Rightarrow \sin 9\theta - \sin \theta = 2\cos \left( {5\theta } \right)\sin \left( {4\theta } \right)$
Now obtained the expression for the L.H.S of the equation as $2\cos \left( {5\theta } \right)\sin \left( {4\theta } \right)$, which is equated to the R.H.S of the equation which is zero, as given below:
$\Rightarrow 2\cos \left( {5\theta } \right)\sin \left( {4\theta } \right) = 0$
Here $\cos \left( {5\theta } \right) = 0$
And $\sin \left( {4\theta } \right) = 0$
Here considering $\cos \left( {5\theta } \right) = 0$, the general solutions of the equation are, as given below:
$\Rightarrow \cos \left( {5\theta } \right) = 0$
$\therefore 5\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}$
$\Rightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{{10}}$
Here n = 0,1,2,3…..
Here $\theta$ should be within the interval $\left[ {0,2\pi } \right]$, hence substituting the values of n till it does not cross the given interval.
The values of $\theta$ are :
$\Rightarrow \dfrac{\pi }{{10}},\dfrac{{3\pi }}{{10}},\dfrac{{5\pi }}{{10}},\dfrac{{7\pi }}{{10}},\dfrac{{9\pi }}{{10}},\dfrac{{11\pi }}{{10}},\dfrac{{13\pi }}{{10}},\dfrac{{15\pi }}{{10}},\dfrac{{17\pi }}{{10}},\dfrac{{19\pi }}{{10}}$.
Here there are 10 solutions of $\theta$.
Now consider $\sin \left( {4\theta } \right) = 0$, the general solutions of the equation are, as given below:
$\Rightarrow \sin \left( {4\theta } \right) = 0$
$\therefore 4\theta = n\pi$
$\Rightarrow \theta = \dfrac{{n\pi }}{4}$
Here n = 0,1,2,3…..
Here $\theta$ should be within the interval $\left[ {0,2\pi } \right]$, hence substituting the values of n till it does not cross the given interval.
The values of $\theta$ are :
$\Rightarrow \dfrac{\pi }{4},\dfrac{{2\pi }}{4},\dfrac{{3\pi }}{4},\dfrac{{4\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{6\pi }}{4},\dfrac{{7\pi }}{4},\dfrac{{8\pi }}{4}$.
Here there are 8 solutions of $\theta$.
$\therefore$In total the solutions of $\theta$ are given by:
$\Rightarrow 10 + 8 = 18$
Hence there are 18 solutions of $\theta$.

Number of solutions of the equation of $\sin 9\theta = \sin \theta$ are 18.

Note:
Here the most important and crucial step is that while substituting the values of n in the general solutions of $\theta$, we have substitute and check in such a way that finally the value of $\theta$ does not cross the interval , that is the value of $\theta$ does not cross $2\pi$. One more important thing is that the formula used here to solve the value of $\theta$ is the trigonometric sum to product formula which is very important.