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Question Number 75763 by mathocean1 last updated on 16/Dec/19

$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{it}\mathrm{in}\mathbb{R}$$$$3\mathrm{cosx}-\sqrt{3}\mathrm{sinx}+\sqrt{6}=0$$

Answered by Ajao yinka last updated on 16/Dec/19

Answered by $@ty@m123 last updated on 17/Dec/19

$$Divideby2\sqrt{3},$$$$\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x+\frac{1}{\sqrt{2}}=0$$$$\sin \frac{\pi }{3}\cos x-\cos \frac{\pi }{3}\sin x=-\frac{1}{\sqrt{2}}$$$$\sin (\frac{\pi }{3}-x)=-\frac{1}{\sqrt{2}}$$$$\sin (x-\frac{\pi }{3})=\frac{1}{\sqrt{2}}$$$$\sin (x-\frac{\pi }{3})=\sin \frac{\pi }{4}$$$$x-\frac{\pi }{3}=n\pi +{(-1)}^n\frac{\pi }{4},n\in \mathbb{Z}$$$$x=n\pi +{(-1)}^n\frac{\pi }{4}+\frac{\pi }{3}$$

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