Question

The number of solutions for the equation $\sin x.\tan 4x=\cos x$ in $\left( 0,\pi \right)$ is\[\]
A.4\[\]
B.7\[\]
C.8\[\]
D. None of these \[\]

Answer 1

Hint: We replace $\tan 4x=\dfrac{\sin 4x}{\cos 4x}$ in the given equation and use the cosine sum of two angles formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ to make equation like $\cos \theta =0$ whose solutions we determine in the form of $\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$. We check how many the solutions lie in the gives interval $\left( 0,\pi \right)$.\[\]

Complete step by step answer:
We know that the solutions of the equation $\cos x=0$ are the odd integral multiples of $\dfrac{\pi }{2}$ that is $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ where $n$ is an integer. We also know the cosine sum of two angle formula where the cosine of sum of two angles say $A$ and $B$ is given by
\[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\]
The given equation is
\[\sin x.\tan 4x=\cos x\]
We know that for any angle $\theta $ , $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. We use it and replace $\tan 4x$ in the above equation to get,
 \[\sin x.\dfrac{\sin 4x}{\cos 4x}=\cos x\]
We multiply $\cos 4x$ in both side of the equation and get
\[\begin{align}
  & \Rightarrow \sin x.\sin 4x=\cos x\cdot \cos 4x \\
 & \Rightarrow \cos 4x\cdot \cos x-\sin 4x\cdot \sin x=0 \\
\end{align}\]
We see that in the above result is in the form of $\cos A\cos B-\sin A\sin B$ where $A=4x$ and $B=x$. We replace left hand side by $\cos \left( A+B \right)$ and get,
\[\begin{align}
  & \Rightarrow \cos \left( 4x+x \right)=0 \\
 & \Rightarrow \cos 5x=0 \\
\end{align}\]

The solutions of the above equations are $5x=\left( 2n+1 \right)\dfrac{\pi }{2}$ or $x=\dfrac{\left( 2n+1 \right)\dfrac{\pi }{2}}{5}=\dfrac{2n\pi }{10}+\dfrac{\pi }{10}$, but we are given that $x$ lies in the interval $\left( 0,\pi \right)$. So let us check for which value of $n$ , $x$ lies within the given interval $\left( 0,\pi \right)$. Let us put $n=0,1,2,3,4,5...$ for $x$ and check how many of them lie in $\left( 0,\pi \right)$.
\[\begin{align}
  & n=0\Rightarrow x=\dfrac{\pi }{10}\in \left( 0,\pi \right) \\
 & n=1\Rightarrow x=\dfrac{3\pi }{10}\in \left( 0,\pi \right) \\
 & n=2\Rightarrow x=\dfrac{5\pi }{10}\in \left( 0,\pi \right) \\
 & n=3\Rightarrow x=\dfrac{7\pi }{10}\in \left( 0,\pi \right) \\
 & n=4\Rightarrow x=\dfrac{9\pi }{10}\in \left( 0,\pi \right) \\
 & n=5\Rightarrow x=\dfrac{11\pi }{10}>\pi \\
\end{align}\]
So the values of $x$ for integer $n>5$ will not lie in the given intervals. Now we put $n=-1,-2,...$ for $n=0,1,2,3,4,5...$ for $x$ and check how many of them lie in $\left( 0,\pi \right)$.
\[n=-1\Rightarrow x=-\pi +\dfrac{\pi }{2}=\dfrac{-\pi }{2}<0\]
So the values of $x$ for integer $n<-1$ will not lie in the given interval $\left( 0,\pi \right)$. Now we have only obtained 5 values for which $x$ lies within the given interval $\left( 0,\pi \right)$ which are $\dfrac{\pi }{10},\dfrac{3\pi }{10},\dfrac{5\pi }{10},\dfrac{7\pi }{10},\dfrac{9\pi }{10} $.

So, the correct answer is “Option D”.

Note: We can also solve the problem by dividing both sides of the equation by $\sin x$, which we can do because $\sin x\ne 0,x\in \left( 0,\pi \right)$. The we find the equation of the form $\tan \theta =\tan \alpha $ solutions are $\theta =n\pi +\alpha ,\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$.