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

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 $\left( {2n + 1} \right)\dfrac{\pi }{2};n \in I$.