Question

Solve the equation $\sqrt{3}\cos x-\sin x=1$

• A. $x=2n\pi -\frac{\pi }{3}\pm \frac{\pi }{3},n\in Z$
• B. $x=2n\pi -\frac{\pi }{6}\pm \frac{\pi }{3},n\in Z$
• C. $x=2n\pi -\frac{\pi }{6}\pm \frac{\pi }{6},n\in Z$
• D. None of these

Answer 1

The correct option is B $x=2n\pi -\frac{\pi }{6}\pm \frac{\pi }{3},n\in Z$

Given equation is $\sqrt{3}\cos x-\sin x=1$

Dividing both sides of the equation by $\sqrt{{\left(\sqrt{3}\right)}^2+{(-1)}^2}=2$,

We obtain $\left(\frac{\sqrt{3}}{2}\right)\cos x-\left(\frac{1}{2}\right)\sin x=\frac{1}{2}$

Observing that $\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$ and $\frac{1}{2}=\sin \frac{\pi }{6}$,

we get $\cos \frac{\pi }{6}\cos x-\sin \frac{\pi }{6}\sin x=\frac{1}{2}$

That is $\cos (x+\frac{\pi }{6})=\frac{1}{2}=\cos \frac{\pi }{3}$,

which gives $x+\frac{\pi }{6}=2n\pi \pm \frac{\pi }{3},n\in Z$

Hence the solution is $x=2n\pi -\frac{\pi }{6}\pm \frac{\pi }{3},n\in Z$