Question

Number of solution(s) of the equation $\dfrac{\sin x}{\cos 3x}+\dfrac{\sin 3x}{\cos 9x}+\dfrac{\sin 9x}{\cos 27x}=0$ in the interval $\left( 0,\dfrac{\pi }{4} \right)$ is:

Answer 1

Hint: First of all, we are going to multiply and divide the first fraction by $\cos x$, second fraction by $\cos 3x$ and third fraction by $\cos 9x$. Then, our expression will look like:$\dfrac{\sin x\cos x}{\cos 3x\cos x}+\dfrac{\sin 3x\cos 3x}{\cos 9x\cos 3x}+\dfrac{\sin 9x\cos 9x}{\cos 27x\cos 9x}=0$ Then we are going to multiply and divide each fraction by 2. After that, we will find that all the numerators are written in the form of $2\sin A\cos A=\sin 2A$. Then we will write the angles in the numerator in the difference of the angles given in the denominator. Then use trigonometric identities to simplify the equation and hence, find the solutions.

Complete step-by-step answer:
We are asked to find the solutions of the following equation:
$\dfrac{\sin x}{\cos 3x}+\dfrac{\sin 3x}{\cos 9x}+\dfrac{\sin 9x}{\cos 27x}=0$
Multiplying and dividing the first, second and third fraction by $\cos x$, $\cos 3x$ and $\cos 9x$ respectively we get,
$\dfrac{\sin x\cos x}{\cos 3x\cos x}+\dfrac{\sin 3x\cos 3x}{\cos 9x\cos 3x}+\dfrac{\sin 9x\cos 9x}{\cos 27x\cos 9x}=0$
Now, multiplying and dividing by 2 on the L.H.S of the above equation we get,
$\dfrac{2\sin x\cos x}{2\cos 3x\cos x}+\dfrac{2\sin 3x\cos 3x}{2\cos 9x\cos 3x}+\dfrac{2\sin 9x\cos 9x}{2\cos 27x\cos 9x}=0$
Now, as you can see that, the numerators of all the fractions are written in the form of $2\sin A\cos A=\sin 2A$ so using this identity in the above equation we get,
$\dfrac{\sin 2x}{2\cos 3x\cos x}+\dfrac{\sin 6x}{2\cos 9x\cos 3x}+\dfrac{\sin 18x}{2\cos 27x\cos 9x}=0$
Taking $\dfrac{1}{2}$ out from the L.H.S of the above equation we get,
$\dfrac{1}{2}\left( \dfrac{\sin 2x}{\cos 3x\cos x}+\dfrac{\sin 6x}{\cos 9x\cos 3x}+\dfrac{\sin 18x}{\cos 27x\cos 9x} \right)=0$
Multiplying 2 on both the sides we get,
$\left( \dfrac{\sin 2x}{\cos 3x\cos x}+\dfrac{\sin 6x}{\cos 9x\cos 3x}+\dfrac{\sin 18x}{\cos 27x\cos 9x} \right)=0$
If you look carefully towards the above equation, you will find that the angles written in the numerators are the difference of the angles written in their corresponding denominators.
$\dfrac{\sin \left( 3x-x \right)}{\cos 3x\cos x}+\dfrac{\sin \left( 9x-3x \right)}{\cos 9x\cos 3x}+\dfrac{\sin \left( 27x-9x \right)}{\cos 27x\cos 9x}=0$……….. Eq. (1)
We know that, there is a trigonometric identity which says that:
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
So, using the above identity in writing the expansion of the numerators of eq. (1) we get:
$\dfrac{\sin 3x\cos x-\cos 3x\sin x}{\cos 3x\cos x}+\dfrac{\sin 9x\cos 3x-\cos 9x\sin 3x}{\cos 9x\cos 3x}+\dfrac{\sin 27x\cos 9x-\cos 27x\sin 9x}{\cos 27x\cos 9x}=0$
Rearranging the above equation we get,
$\begin{align}
  & \dfrac{\sin 3x\cos x}{\cos 3x\cos x}-\dfrac{\cos 3x\sin x}{\cos 3x\cos x}+\dfrac{\sin 9x\cos 3x}{\cos 9x\cos 3x}-\dfrac{\cos 9x\sin 3x}{\cos 9x\cos 3x}+\dfrac{\sin 27x\cos 9x}{\cos 27x\cos 9x}-\dfrac{\cos 27x\sin 9x}{\cos 27x\cos 9x}=0 \\
 & \Rightarrow \tan 3x-\tan x+\tan 9x-\tan 3x+\tan 27x-\tan 9x=0 \\
\end{align}$
In the above equation, as you can see that positive $\left( \tan 3x,\tan 9x \right)$ will be cancelled out with negative $\left( \tan 3x,\tan 9x \right)$ and we get,
$\begin{align}
  & \tan 27x-\tan 9x=0 \\
 & \Rightarrow \tan 27x=\tan 9x......Eq.(2) \\
\end{align}$
Now, we know that if:
$\begin{align}
  & \tan \theta =\tan \alpha \\
 & \Rightarrow \theta =n\pi +\alpha \\
\end{align}$
Using the above relation in eq. (2) we get,
$\begin{align}
  & 27x=n\pi +x \\
 & \Rightarrow 26x=n\pi \\
 & \Rightarrow x=\dfrac{n\pi }{26} \\
\end{align}$
Now, it is given in the above problem that we have to write solutions between $\left( 0,\dfrac{\pi }{4} \right)$ so we are substituting different values in “n” right from n equals 1 till the value of x should be less than $\dfrac{\pi }{4}$.
$\dfrac{1\pi }{26},\dfrac{2\pi }{26},\dfrac{3\pi }{26},\dfrac{4\pi }{26},\dfrac{5\pi }{26},\dfrac{6\pi }{26}$
If we substitute n equals 7 then we get $\dfrac{7\pi }{26}$ which is greater than $\dfrac{\pi }{4}$. Hence, we have stopped till n equals 6.
Hence, there are 6 solutions for the given equation.

Note: The plausible mistake that could happen in the above solution is that you might have been considered the extreme values 0 and $\dfrac{\pi }{4}$ in the interval $\left( 0,\dfrac{\pi }{4} \right)$ then you will get 0 also as a solution and your total number of solutions then became 7. This is the wrong answer because we cannot consider the extreme values of the interval because the interval is written with open brackets and in open brackets, extreme values are excluded.