Question

The number of solutions of the equation $ \sin x\cos 3x = \sin 3x\cos 5x $ in $ \left[ {0,\dfrac{\pi }{2}} \right] $ is:
(A) $ 3 $
(B) $ 4 $
(C) $ 5 $
(D) $ 6 $

Answer 1

Hint: The given question involves solving of a trigonometric equation and finding value of angle x that satisfy the given equation and lie in the range of $ \left[ {0,\dfrac{\pi }{2}} \right] $ . There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities. We will use the trigonometric formulae $ 2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right) $ to simplify the equation at first.

Complete step-by-step answer:
In the given problem, we have to solve the trigonometric equation $ \sin x\cos 3x = \sin 3x\cos 5x $ and find the values of x that satisfy the given equation and lie in the range of $ \left[ {0,\dfrac{\pi }{2}} \right] $ .
So, In order to solve the given trigonometric equation $ \sin x\cos 3x = \sin 3x\cos 5x $ , we should first take all the terms to the left side of the equation.
Transposing all the terms to left side of the equation, we get,
 $ \Rightarrow \sin x\cos 3x - \sin 3x\cos 5x = 0 $
Now, we multiply all the terms of the equation by two. So, we get,
 $ \Rightarrow 2\sin x\cos 3x - 2\sin 3x\cos 5x = 0 $
Now, we can use the trigonometric formula $ 2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right) $ in the equation. So, we get,
\[ \Rightarrow \left[ {\sin \left( {x + 3x} \right) + \sin \left( {x - 3x} \right)} \right] - \left[ {\sin \left( {3x + 5x} \right) + \sin \left( {3x - 5x} \right)} \right] = 0\]
Simplifying the equation, we get,
\[ \Rightarrow \sin 4x + \sin \left( { - 2x} \right) - \sin 8x - \sin \left( { - 2x} \right) = 0\]
Cancelling the like terms with opposite signs, we get,
\[ \Rightarrow \sin 4x - \sin 8x = 0\]
Now, using the double angle formula for sine $ \sin 2\theta = 2\sin \theta \cos \theta $ , we get,
\[ \Rightarrow \sin 4x - 2\sin 4x\cos 4x = 0\]
\[ \Rightarrow \sin 4x\left( {1 - 2\cos 4x} \right) = 0\]
Now, if the product of two terms is zero, either of the terms can be zero. So, we get,
Either $ \sin 4x = 0 $ or \[\left( {1 - 2\cos 4x} \right) = 0\]
 $ \Rightarrow \sin 4x = 0 $ or $ \Rightarrow \cos 4x = \cos \left( {\dfrac{\pi }{3}} \right) $
So, we know that the sine function is equal to zero for multiples of $ \pi $ .
Hence, $ 4x = n\pi $
 $ \Rightarrow x = \left( {\dfrac{{n\pi }}{4}} \right) $
Here, for $ n = 0 $ ,
 $ \Rightarrow x = 0 $
Now, for $ n = 1 $ ,
 $ \Rightarrow x = \left( {\dfrac{\pi }{4}} \right) $
Also, for $ n = 2 $ ,
 $ \Rightarrow x = \left( {\dfrac{\pi }{2}} \right) $
So, the values of x satisfying the trigonometric equation $ \sin 4x = 0 $ are: $ 0,\dfrac{\pi }{4},\dfrac{\pi }{2} $ .
Now, $ \cos 4x = \cos \left( {\dfrac{\pi }{3}} \right) $
We know the general equation for the equation of form $ \cos x = \cos y $ is $ x = 2n\pi \pm y $ .
Hence, $ 4x = 2n\pi \pm \dfrac{\pi }{3} $
Putting in the value of n as zero. We get,
 $ \Rightarrow 4x = 2\left( 0 \right)\pi \pm \dfrac{\pi }{3} $
 $ \Rightarrow x = \pm \dfrac{\pi }{{12}} $
So, the value of x as $ \dfrac{\pi }{{12}} $ lies in $ \left[ {0,\dfrac{\pi }{2}} \right] $ .
Hence, the values of x that satisfy the original equation $ \sin x\cos 3x = \sin 3x\cos 5x $ are: \[0,\dfrac{\pi }{{12}},\dfrac{\pi }{4},\dfrac{\pi }{2}\]. Hence, there are four solutions to the equation $ \sin x\cos 3x = \sin 3x\cos 5x $ in $ \left[ {0,\dfrac{\pi }{2}} \right] $
So, the correct answer is “Option B”.

Note: We must take care of the calculations and verify them once so as to be sure of the answer. We should know the trigonometric formulas such as the double angle formula of sine so as to solve the given problem. We must remember the formats of general solutions of some standard trigonometric equations such as $ \cos x = \cos y $ .