The general solution of $4 \cos 2 x-4 \sqrt{3} \sin 2 x+\cos 3 x-$ $\sqrt{\sin 3 x+\cos x-\sqrt{3} \sin x}=0$

The general solution of $4 \cos 2 x-4 \sqrt{3} \sin 2 x+\cos 3 x-$ $\sqrt{\sin 3 x+\cos x-\sqrt{3} \sin x}=0$
  1. $\frac{n \pi}{2}-\frac{\pi}{3}$
  2. $\frac{n \pi}{2}+\frac{\pi}{6}$
  3. $\frac{n \pi}{2}+\frac{\pi}{12}$
  4. $\frac{n \pi}{2}-\frac{\pi}{12}$

Solution

Given, $4 \cos 2 x-4 \sqrt{3} \sin 2 x+\cos 3 x-\sqrt{3} \sin 3 x +\cos x-\sqrt{3} \sin x=0$ $\Rightarrow 4 \cos 2 x-4 \sqrt{3} \sin 2 x+(\cos 3 x+\cos x)-\sqrt{3}(\sin 3 x+\sin x)=0$ $\Rightarrow 4 \cos 2 x-4 \sqrt{3} \sin 2 x+2 \cos 2 x \cdot \cos x-2 \sqrt{3} \sin 2 x \cos x=0$ $\begin{aligned} & \Rightarrow(2+\cos x)(2 \cos 2 x-2 \sqrt{3} \sin 2 x)=0 \\ & \Rightarrow \cos 2 x=\sqrt{3} \sin 2 x \Rightarrow \tan 2 x=\frac{1}{\sqrt{3}} \\ & \Rightarrow 2 x=\frac{\pi}{6}+n \pi \Rightarrow x=\frac{\pi}{12}+\frac{n \pi}{2}\end{aligned}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

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