Question

The number of solution of the equation \[\sin \left( {9x} \right) + \sin \left( {3x} \right) = 0\]in the closed interval \[\left[ {0,2\pi } \right]\] is
A) 7
B) 13
C) 19
D) 25

Answer 1

Hint: In this question first use the formula for \[\sin 3x\]to simplify sin 9x in the given equation after that solve the equation in order to get the different values of x in the given range.

Complete step by step solution:
\[\sin \left( {9x} \right) + \sin \left( {3x} \right) = 0\],\[x \in \left[ {0,2\pi } \right]\] (1)
Now we use the formula of \[\sin 3x = 3\sin x - 4{\sin ^3}x\],
On sin 9x we get, \[\sin 9x = 3\sin 3x - 4{\sin ^3}3x\],
Substituting the value of sin9x in (1) we get,
\[ \Rightarrow 3\sin 3x - 4{\sin ^3}3x + \sin 3x = 0\]
\[ \Rightarrow 4\sin 3x - 4{\sin ^3}3x = 0\]
Taking 4 sin 3x common, we get,
\[ \Rightarrow 4\sin 3x\left( {1 - {{\sin }^2}3x} \right) = 0\]
Now we have,
\[ \Rightarrow 4\sin 3x = 0\]Or \[\left( {1 - {{\sin }^2}3x} \right) = 0\]
\[ \Rightarrow 4\sin 3x = 0\] Or \[{\cos ^2}x\]= 0 \[ \ldots \left[
  \because {\cos ^2}x + {\sin ^2}x = 1 \\
   \Rightarrow 1 - {\sin ^2}x = {\cos ^2}x \\
   \right]\]
\[ \Rightarrow 4\sin 3x = 0\] Or \[\cos x = 0\]
\[ \Rightarrow 3x = n\pi ,n \in I\]Or \[3x = k\pi + \dfrac{\pi }{2},k \in I\]
\[ \ldots \] [zero of a sine function is \[n\pi ;n \in I\]and zeros of cosine function is \[\left( {k\pi + 1} \right)\dfrac{\pi }{2};k \in I\] ]
\[ \Rightarrow x = \dfrac{{n\pi }}{3},n \in I\]Or \[x = \dfrac{{k\pi }}{3} + \dfrac{\pi }{6},k \in I\]
\[ \Rightarrow x = \dfrac{{n\pi }}{3},n \in I\]Or \[x = \dfrac{{2k\pi }}{6} + \dfrac{\pi }{6},k \in I\]
First consider \[x = \dfrac{{n\pi }}{3},n \in I\]
For n = 0 we have
\[x = 0\]
For n=1
\[x = \dfrac{{\left( 1 \right)\pi }}{3} = \dfrac{\pi }{3}\]
For n=2
\[x = \dfrac{{\left( 2 \right)\pi }}{3} = \dfrac{{2\pi }}{3}\]
For n=3
\[x = \dfrac{{\left( 3 \right)\pi }}{3} = \pi \]
For n=4
\[x = \dfrac{{\left( 4 \right)\pi }}{3} = \dfrac{{4\pi }}{3}\]
For n=5
\[x = \dfrac{{\left( 5 \right)\pi }}{3} = \dfrac{{5\pi }}{3}\]
For n=6
\[x = \dfrac{{\left( 6 \right)\pi }}{3} = 2\pi \]
Hence, we have
\[x = 0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3},2\pi \]\[ \ldots \left[ 1 \right]\]
Now, consider \[x = \dfrac{{2k\pi }}{6} + \dfrac{\pi }{6},k \in I\]
For k=0
\[x = \dfrac{{2\left( 0 \right)\pi }}{6} + \dfrac{\pi }{6}\]
\[ \Rightarrow x = \dfrac{\pi }{6}\]
For k=1
\[x = \dfrac{{2\left( 1 \right)\pi }}{6} + \dfrac{\pi }{6} = \dfrac{{3\pi }}{6}\]
\[ \Rightarrow x = \dfrac{\pi }{2}\]
For k=2
\[x = \dfrac{{2\left( 2 \right)\pi }}{6} + \dfrac{\pi }{6} = \dfrac{{4\pi }}{6} + \dfrac{\pi }{6}\]
\[ \Rightarrow x = \dfrac{{5\pi }}{6}\]
For k=3
\[x = \dfrac{{2\left( 3 \right)\pi }}{6} + \dfrac{\pi }{6} = \dfrac{{6\pi }}{6} + \dfrac{\pi }{6}\]
\[ \Rightarrow x = \dfrac{{7\pi }}{6}\]
For k=4
\[x = \dfrac{{2\left( 4 \right)\pi }}{6} + \dfrac{\pi }{6} = \dfrac{{8\pi }}{6} + \dfrac{\pi }{6}\]
\[ \Rightarrow x = \dfrac{{9\pi }}{6}\]
\[ \Rightarrow x = \dfrac{{3\pi }}{2}\]
For k=5
\[x = \dfrac{{2\left( 5 \right)\pi }}{6} + \dfrac{\pi }{6} = \dfrac{{10\pi }}{6} + \dfrac{\pi }{6}\]
\[ \Rightarrow x = \dfrac{{11\pi }}{6}\]
Hence, we have
\[x = \dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}\]\[ \ldots \left[ 2 \right]\]
From 1 and 2 we get the values of x are as follows:
\[\left\{ {0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3},2\pi ,\dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}} \right\}\]
So the total solutions are \[\left\{ {0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3},2\pi ,\dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}} \right\}\]
Thus, the total numbers of solution are 13
Hence, option B. 13 is the correct answer.

Note: sine and cosine function are periodic with a period of \[2\pi \]. The domain of each function is \[\left( { - \infty ,\infty } \right)\]and range is \[\left[ { - 1,1} \right]\]. The zeros of sine function are \[n\pi ;n \in I\] and zeros of cosine function are \[\left( {2n + 1} \right)\dfrac{\pi }{2};n \in I\].