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Question: The number of solutions for the equation \(\sin x.\tan 4x=\cos x\) in \(\left( 0,\pi \right)\) is\[\...

The number of solutions for the equation sinx.tan4x=cosx\sin x.\tan 4x=\cos x in (0,π)\left( 0,\pi \right) isA.4 A.4
B.7C.8 C.8
D. None of these $$$$

Explanation

Solution

We replace tan4x=sin4xcos4x\tan 4x=\dfrac{\sin 4x}{\cos 4x} in the given equation and use the cosine sum of two angles formula cos(A+B)=cosAcosBsinAsinB\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B to make equation like cosθ=0\cos \theta =0 whose solutions we determine in the form of θ=(2n+1)π2\theta =\left( 2n+1 \right)\dfrac{\pi }{2}. We check how many the solutions lie in the gives interval (0,π)\left( 0,\pi \right).$$$$

Complete step by step answer:
We know that the solutions of the equation cosx=0\cos x=0 are the odd integral multiples of π2\dfrac{\pi }{2} that is x=(2n+1)π2x=\left( 2n+1 \right)\dfrac{\pi }{2} where nn is an integer. We also know the cosine sum of two angle formula where the cosine of sum of two angles say AA and BB is given by
cos(A+B)=cosAcosBsinAsinB\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B
The given equation is
sinx.tan4x=cosx\sin x.\tan 4x=\cos x
We know that for any angle θ\theta , tanθ=sinθcosθ\tan \theta =\dfrac{\sin \theta }{\cos \theta }. We use it and replace tan4x\tan 4x in the above equation to get,
sinx.sin4xcos4x=cosx\sin x.\dfrac{\sin 4x}{\cos 4x}=\cos x
We multiply cos4x\cos 4x in both side of the equation and get

& \Rightarrow \sin x.\sin 4x=\cos x\cdot \cos 4x \\\ & \Rightarrow \cos 4x\cdot \cos x-\sin 4x\cdot \sin x=0 \\\ \end{aligned}$$ We see that in the above result is in the form of $\cos A\cos B-\sin A\sin B$ where $A=4x$ and $B=x$. We replace left hand side by $\cos \left( A+B \right)$ and get, $$\begin{aligned} & \Rightarrow \cos \left( 4x+x \right)=0 \\\ & \Rightarrow \cos 5x=0 \\\ \end{aligned}$$ The solutions of the above equations are $5x=\left( 2n+1 \right)\dfrac{\pi }{2}$ or $x=\dfrac{\left( 2n+1 \right)\dfrac{\pi }{2}}{5}=\dfrac{2n\pi }{10}+\dfrac{\pi }{10}$, but we are given that $x$ lies in the interval $\left( 0,\pi \right)$. So let us check for which value of $n$ , $x$ lies within the given interval $\left( 0,\pi \right)$. Let us put $n=0,1,2,3,4,5...$ for $x$ and check how many of them lie in $\left( 0,\pi \right)$. $$\begin{aligned} & 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{aligned}$$ So the values of $x$ for integer $n>5$ will not lie in the given intervals. Now we put $n=-1,-2,...$ for $n=0,1,2,3,4,5...$ for $x$ and check how many of them lie in $\left( 0,\pi \right)$. $$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 $\theta =n\pi +\alpha ,\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$.