Question

The general solution of $\sin x–3\sin 2x+\sin 3x=\cos x–3\cos 2x+\cos 3x$ is

• A. $n\pi +\frac{\pi }{8}$
• B. $n\frac{\pi }{2}+\frac{\pi }{8}$
• C. $(-1{)}^n\frac{n\pi }{2}+\frac{\pi }{8}$
• D. $2n\pi +co{s}^{-1}\frac{3}{2}$

Answer 1

The correct option is B

$n\frac{\pi }{2}+\frac{\pi }{8}$

Find the general solution of the given expression

Given, $\sin x–3\sin 2x+\sin 3x=\cos x–3\cos 2x+\cos 3x$

$$\begin{gathered} \Rightarrow (s\mathrm{in}x+\sin 3x)-3\sin 2x=(\cos x+\cos 3x)-3\cos 2x \\ \Rightarrow 2sin2x\cos x-3\sin 2x=2\cos 2x\cos x-3cos2x\left[\because \sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\&\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\right] \\ \Rightarrow \sin 2x(2\cos x-3)=\cos 2x(2\cos x-3) \\ \Rightarrow \sin 2x=\cos 2x \\ \Rightarrow \frac{\sin 2x}{\cos 2x}=1 \\ \Rightarrow \tan 2x=1\left[\because \frac{\sin \theta }{\cos \theta }=\tan \theta \right] \\ \Rightarrow \tan 2x=\tan \frac{\mathrm{\pi }}{4}\left[\because \tan \frac{\mathrm{\pi }}{4}=1\right] \\ \Rightarrow 2x=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=\mathrm{n}\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{8} \end{gathered}$$

Hence, option (B) $n\frac{\pi }{2}+\frac{\pi }{8}$, is the correct answer.