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Question Number 121303 by talminator2856791 last updated on 06/Nov/20

$$$$$$$$$$$$$$$$$$\cos x-\sqrt{3}\sin x=2\cos 2x$$$$$$$$$$

Commented by liberty last updated on 06/Nov/20

$$\Rightarrow \cos \mathrm{x}-\sqrt{3}\sin \mathrm{x}=\mathrm{k}\cos (\mathrm{x}-\alpha )$$$$\mathrm{k}=\sqrt{1^2+{(-\sqrt{3})}^2}=2$$$$\tan \alpha =\frac{-\sqrt{3}}{1}\left(4^{\mathrm{th}}\mathrm{quadrant}\right)$$$$\alpha =300°$$$$\Rightarrow 2\cos (\mathrm{x}-300°)=2\cos 2\mathrm{x}$$$$\Rightarrow \cos 2\mathrm{x}=\cos (\mathrm{x}-300°)$$$$$$

Answered by Bird last updated on 06/Nov/20

$$cosx-\sqrt{3}sinx=2cos\left(2x\right)\Rightarrow$$$$2(\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx)=2cos\left(2x\right)$$$$\Rightarrow cosxcos\left(\frac{\pi }{3}\right)-sinxsin\left(\frac{\pi }{3}\right)=cos\left(2x\right)$$$$\Rightarrow cos(x+\frac{\pi }{3})=cos\left(2x\right)\Rightarrow$$$$x+\frac{\pi }{3}=2x+2k\pi orx+\frac{\pi }{3}=-2x+2k\pi$$$$\Rightarrow -x=-\frac{\pi }{3}+2k\pi or3x=-\frac{\pi }{3}+2k\pi \Rightarrow$$$$x=\frac{\pi }{3}-2k\pi orx=-\frac{\pi }{9}+\frac{2k\pi }{3}$$$$k\in Z$$$$$$

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