Question

# The solution of the equation $\sin x + 3\sin 2x + \sin 3x = \cos x + 3\cos 2x + \cos 3x$ in the interval $0 \leqslant x \leqslant 2\pi$ are$(a){\text{ }}\dfrac{\pi }{3},\dfrac{{5\pi }}{8},\dfrac{{2\pi }}{3} \\ (b){\text{ }}\dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{8},\dfrac{{13\pi }}{8} \\ (c){\text{ }}\dfrac{{4\pi }}{3},\dfrac{{9\pi }}{3},\dfrac{{2\pi }}{3},\dfrac{{13\pi }}{8} \\ (d){\text{ }}\dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{3},\dfrac{{4\pi }}{3} \\$

Hint â€“ In this question we have to find the solution of the given equation, use trigonometric identities like $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ to simplify for the value of x, but keep in mind that the solution is asked to be found only for the interval of $0 \leqslant x \leqslant 2\pi$.

Given equation is
$\sin x + 3\sin 2x + \sin 3x = \cos x + 3\cos 2x + \cos 3x$
$\Rightarrow \sin x + \sin 3x + 3\sin 2x = \cos x + \cos 3x + 3\cos 2x$
We have to find out the solution of this equation in the interval$0 \leqslant x \leqslant 2\pi$.
Now as we know
$\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\ \cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\$
So, use this property in the above equation we have,
$\Rightarrow \sin x + \sin 3x + 3\sin 2x = \cos x + \cos 3x + 3\cos 2x$
$\Rightarrow 2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + 3\sin 2x = 2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + 3\cos 2x$
Now simplify the above equation we have,
$\Rightarrow 2\sin 2x\cos x + 3\sin 2x = 2\cos 2x\cos x + 3\cos 2x$ $\left[ {\because \cos \left( { - \theta } \right) = \cos \theta } \right]$
Now take common the common terms we have,
$\Rightarrow \sin 2x\left( {2\cos x + 3} \right) = \cos 2x\left( {2\cos x + 3} \right)$
$\Rightarrow \left( {2\cos x + 3} \right)\left( {\sin 2x - \cos 2x} \right) = 0$â€¦â€¦â€¦â€¦â€¦â€¦. (1)
$\Rightarrow 2\cos x + 3 = 0$
$\Rightarrow \cos x = \dfrac{{ - 3}}{2}$ (Which is not possible as the value of cosine is always lie between -1 to 1)
Again from equation (1) we have,
$\Rightarrow \left( {\sin 2x - \cos 2x} \right) = 0$
$\Rightarrow \sin 2x = \cos 2x$
$\Rightarrow \dfrac{{\sin 2x}}{{\cos 2x}} = 1$
$\Rightarrow \tan 2x = 1 = \tan \left( {n\pi + \dfrac{\pi }{4}} \right)$ (General solution where n = 0, 1, 2, 3, â€¦â€¦â€¦â€¦â€¦â€¦â€¦. )
So on comparing
$\Rightarrow 2x = n\pi + \dfrac{\pi }{4}$
$\Rightarrow x = n\dfrac{\pi }{2} + \dfrac{\pi }{8}$
Now, we have to find out the value of x in the interval$0 \leqslant x \leqslant 2\pi$.
For n = 0
$\Rightarrow x = \dfrac{\pi }{8}$.
For n = 1
$\Rightarrow x = \dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{5\pi }}{8}$
For n = 2
$\Rightarrow x = 2\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{9\pi }}{8}$
For n = 3
$\Rightarrow x = 3\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{13\pi }}{8}$
For n = 4
$\Rightarrow x = 4\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{17\pi }}{8} > 2\pi$ (So this solution is not considered)
Hence the required solution is $x = \dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{8},\dfrac{{13\pi }}{8}$
Hence option (b) is correct.

Note: Whenever we face such types of problems the key concept is firstly to reach to the value of x but application of various trigonometric and algebraic identities. As the trigonometric functions involved are mostly periodic in nature therefore the solution also varies in different intervals so we need to keep in mind about the interval in which the solution is being asked.