Question

The general solution of $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ is
A.$n\pi +\dfrac{\pi }{8}$
B.$\dfrac{n\pi }{2}+\dfrac{\pi }{8}$
C.${{\left( -1 \right)}^{n}}\left( \dfrac{n\pi }{2}+\dfrac{\pi }{8} \right)$
D.$2n\pi -\dfrac{\pi }{8}$

Answer 1

Hint: We know the formula, $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ . Use this formula and simplify $\sin x+\sin 3x$ and $\cos x+\cos 3x$ and then, substitute it in the equation $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ . We know that the maximum value of cosine function is 1. We know that $\tan \dfrac{\pi }{4}=1$ and the general solution of $\tan x=\tan \dfrac{\pi }{4}$ is $x=n\pi +\dfrac{\pi }{4}$ .

Complete step-by-step answer:
According to the question, we have to find the general solution of,
$\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ ……………..(1)
We know the formula,
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ …………………(2)
Replacing A by x and B by 3x in the equation (2), we get
$\sin x+\sin 3x=2\sin \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)$
$\Rightarrow \sin x+\sin 3x=2\sin 2x\cos (-x)$ ……………….(4)
We know the property that, $\cos (-x)=\cos x$ .
Using this property in equation (4), we get
$\Rightarrow \sin x+\sin 3x=2\sin 2x\cos (-x)$
$\Rightarrow \sin x+\sin 3x=2\sin 2x\cos x$ …………………….(5)
We also know the formula,
$\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ …………………(6)
Replacing A by x and B by 3x in the equation (6), we get
$\cos x+\cos 3x=2\cos \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)$
$\Rightarrow \cos x+\cos 3x=2\cos 2x\cos (-x)$ ……………….(7)
We know the property that, $\cos (-x)=\cos x$ .
Using this property in equation (7), we get
$\Rightarrow \cos x+\cos 3x=2\cos 2x\cos (-x)$
$\Rightarrow \cos x+\cos 3x=2\cos 2x\cos x$ …………………….(8)
From equation (1), we have
$\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$
$\Rightarrow \sin x+\sin 3x-3\sin 2x=\cos x+\cos 2x-3\cos 2x$ …………………(9)
Now, from equation (5), equation (8), and equation (9), we get
$\Rightarrow \sin x+\sin 3x-3\sin 2x=\cos x+\cos 3x-3\cos 2x$
$\Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos 2x\cos x-3\cos 2x$
Taking $\sin 2x$ in the above equation, we get
$\Rightarrow \sin 2x\left( 2\cos x-3 \right)=\cos 2x\left( 2\cos x-3 \right)$

& \Rightarrow \sin 2x\left( 2\cos x-3 \right)-\cos 2x\left( 2\cos x-3 \right)=0 \\\ & \Rightarrow \left( 2\cos x-3 \right)\left( \sin 2x-\cos 2x \right)=0 \\\ \end{aligned}$$ $$\left( 2\cos x-3 \right)=0$$ or $$\left( \sin 2x-\cos 2x \right)=0$$ . First, let us solve for $$\left( 2\cos x-3 \right)=0$$ $$\left( 2\cos x-3 \right)=0$$ $$\begin{aligned} & \Rightarrow 2\cos x=3 \\\ & \Rightarrow \cos x=\dfrac{3}{2} \\\ \end{aligned}$$ We know that the maximum value of the cosine function is 1 and here, we have $$\cos x=\dfrac{3}{2}$$ , which is not possible for any value of x. Now, we have to solve $$\left( \sin 2x-\cos 2x \right)=0$$ . $$\left( \sin 2x-\cos 2x \right)=0$$ $$\Rightarrow \sin 2x=\cos 2x$$ $$\Rightarrow \dfrac{\sin 2x}{\cos 2x}=1$$ …………….(10) We know that, $$\dfrac{\sin A}{\cos A}=\tan A$$ . Replacing A by 2x in $$\dfrac{\sin A}{\cos A}=\tan A$$ , we get $$\dfrac{\sin 2x}{\cos 2x}=\tan 2x$$ ………(11) From equation (10) and equation (11), we get $$\Rightarrow \dfrac{\sin 2x}{\cos 2x}=1$$ $$\Rightarrow \tan 2x=1$$ …………..(12) Using $$\tan \dfrac{\pi }{4}=1$$ in equation (12), we get $$\begin{aligned} & \Rightarrow \tan 2x=\tan \dfrac{\pi }{4} \\\ & \Rightarrow 2x=\dfrac{\pi }{4} \\\ \end{aligned}$$ Using $$\tan \dfrac{5\pi }{4}=1$$ in equation (12), we get $$\begin{aligned} & \Rightarrow \tan 2x=\tan \dfrac{5\pi }{4} \\\ & \Rightarrow 2x=\dfrac{5\pi }{4} \\\ \end{aligned}$$ We have $$2x=\dfrac{\pi }{4}$$ , $$2x=\dfrac{5\pi }{4}=\pi +\dfrac{\pi }{4}$$ ………………. Now, in general we can say that, $$\begin{aligned} & 2x=n\pi +\dfrac{\pi }{4} \\\ & \Rightarrow x=\dfrac{n\pi }{2}+\dfrac{\pi }{8} \\\ \end{aligned}$$ So, the correct option is option (B). Note: In this question, one might think to find the general solution of $$\cos x=\dfrac{3}{2}$$ . But we don’t have any solution for $$\cos x=\dfrac{3}{2}$$ because the maximum value of cosine function is 1. Also, one can think to use formula $$\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$$ and $$\sin 2x=2\sin x\cos x$$ in the equation $$\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$$ . If we do so, then our equation will become complex to solve.