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

The general solution of $\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x$ is
  1. $x=\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
  2. $x=2 \mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
  3. $x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
  4. $x=\frac{\mathrm{n} \pi}{2}+\frac{\pi}{8}, \mathrm{n} \in \mathbb{Z}$

Solution

$\begin{aligned} & \sin x-3 \sin 2 x+\sin 3 x \\ & =\cos x-3 \cos 2 x+\cos 3 x \\ & \Rightarrow(\sin x+\sin 3 x)-3 \sin 2 x-(\cos x+\cos 3 x) \\ & +3 \cos 2 x=0 \\ & \Rightarrow 2 \sin 2 x \cos x-3 \sin 2 x-2 \cos 2 x \cos x \\ & +3 \cos 2 x=0 \\ & \Rightarrow \sin 2 x(2 \cos x-3)-\cos 2 x(2 \cos x-3)=0 \\ & \Rightarrow(\sin 2 x-\cos 2 x)(2 \cos x-3)=0 \\ & \Rightarrow \cos 2 x=\sin 2 x \quad \ldots\left[\because \cos x \neq \frac{3}{2}\right] \end{aligned}$ $\begin{aligned} & \Rightarrow \cos 2 x=\cos \left(\frac{\pi}{2}-2 x\right) \\ & \Rightarrow 2 x=2 n \pi \pm\left(\frac{\pi}{2}-2 x\right) \end{aligned}$
Neglecting $(-)$ sign, we get $x=\frac{\mathrm{n} \pi}{2}+\frac{\pi}{8}$

Asked in: MHT CET 2024 (09 May Shift 1)

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