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Solve \(\sqrt{3}\) cosx − sinx =1

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We have: \(\sqrt{3 }\)cos x − sin x = 1 

Dividing both sides by 2, we get 

cos x. \(\frac{\sqrt{3}}{2}\) − sin x . \(\frac{1}{2}\) = \(\frac{1}{2}\) 

⇒ cos x . cos \(\frac{\pi}{6}\) − sin x . sin \(\frac{\pi}{6}\) = \(\frac{1}{2}\) 

⇒ cos (x + \(\frac{\pi}{6}\) ) = cos ( \(\frac{\pi}{3}\)

⇒ x + \(\frac{\pi}{6}\) = \(2n\pi\) ± \(\frac{\pi}{3}\)

⇒ x = \(2nx\) ± \(\frac{\pi}{3}\)\(\frac{\pi}{6}\) 

⇒ x = \(2n\pi\) + \(\frac{\pi}{3}\)\(\frac{\pi}{6}\) or x = \(2nx\)\(\frac{\pi}{3}\) −   \(\frac{\pi}{6}\)  

⇒ x = \(2n\pi\) + \(\frac{\pi}{6}\) or x = \(2n\pi\)\(\frac{\pi}{2}\)

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