Question

Total number of solutions to the equation $\sin x \times \tan 4x = \cos x$ belonging to $\left( {0,\pi } \right)$ are:
A) 4
B) 7
C) 8
D) 5

Answer 1

We can write the tan function in terms of sine and cosine. Then we can cross multiply and rearrange the equation. Then we can simplify the equation using trigonometric identities. Then we can solve for x to find the general solution and find all the solutions in the given interval. Then the number of solutions in the interval is the required solution.

Complete step by step solution:
We have the trigonometric equation $\sin x \times \tan 4x = \cos x$
We know that $\tan A = \dfrac{{\sin A}}{{\cos A}}$ . On substituting this in the equation, we get,
$\Rightarrow \sin x.\dfrac{{\sin 4x}}{{\cos 4x}} = \cos x$
On multiplying throughout with $\cos 4x$ , we get,
$\Rightarrow \sin x\sin 4x = \cos x\cos 4x$
On rearranging, we get,
$\Rightarrow \cos x\cos 4x - \sin x\sin 4x = 0$
The LHS is of the form $\cos A\cos B - \sin A\sin B$ . We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ .
$\Rightarrow \cos \left( {x + 4x} \right) = 0$
Hence, we have,
$\Rightarrow \cos \left( {5x} \right) = 0$ … (1)
We know that cos function is zero for odd multiples of $\dfrac{\pi }{2}$
$\Rightarrow \cos \dfrac{{\left( {2n + 1} \right)\pi }}{2} = 0$ …. (2) where n is any integer
On equating (1) and (2), we get,
$\Rightarrow \cos \left( {5x} \right) = \cos \dfrac{{\left( {2n + 1} \right)\pi }}{2}$
Now we can equate the angles.
$\Rightarrow \dfrac{{\left( {2n + 1} \right)\pi }}{2} = 5x$
On dividing throughout with 5, we get,
$\Rightarrow x = \dfrac{{\left( {2n + 1} \right)\pi }}{{10}}$
Now we can give values to n and find the corresponding value of x.
When n is 0, $x = \dfrac{{\left( {2 \times 0 + 1} \right)\pi }}{{10}}$
$\Rightarrow x = \dfrac{\pi }{{10}}$
When n is 1, $x = \dfrac{{\left( {2 \times 1 + 1} \right)\pi }}{{10}}$
$\Rightarrow x = \dfrac{{3\pi }}{{10}}$
When n is 2, $x = \dfrac{{\left( {2 \times 2 + 1} \right)\pi }}{{10}}$
$\Rightarrow x = \dfrac{{5\pi }}{{10}}$
When n is 3, $x = \dfrac{{\left( {2 \times 3 + 1} \right)\pi }}{{10}}$
$\Rightarrow x = \dfrac{{7\pi }}{{10}}$
When n is 4, $x = \dfrac{{\left( {2 \times 4 + 1} \right)\pi }}{{10}}$
$\Rightarrow x = \dfrac{{9\pi }}{{10}}$
For all other values of n, the value of x goes beyond the interval $\left( {0,\pi } \right)$ . Therefore, there are 5 solutions of x in the interval.
Therefore, the required number of solutions is 5.

So, the correct answer is option D.

Note:
For solving any variable, we must take all the terms with the variable to one side and make the other side zero. The trigonometric equations used in this problem are, $\tan A = \dfrac{{\sin A}}{{\cos A}}$ and $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ . We must know that cos is zero for angles that are odd multiples of $\dfrac{\pi }{2}$ . We took n as integers. We did not take the negative integers as for negative values of n, x will become negative which is not in the given interval. For n greater than 5 also, we get the values of x which are not in the required interval.