Question

Solve : $\sqrt{3}cosx-sinx=1.$

Answer 1

Given: $\sqrt{3}cosx-sinx=\mathrm{1.....}\left(i\right)$

Dividing both sides of (i)by $\sqrt{(\sqrt{3}{)}^2+(-1{)}^2,}$ i.e., by 2, we get

$\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx=\frac{1}{2}$

$\Rightarrow cosxcos\frac{\pi }{6}-sinxsin\frac{\pi }{6}=\frac{1}{2}$

$\Rightarrow cos(x+\frac{\pi }{6})=cos\frac{\pi }{3}[\because \frac{1}{2}=cos\frac{\pi }{3}]$

$\Rightarrow (x+\frac{\pi }{6})=2n\pi \pm \frac{\pi }{3}$, where $n\in I[\because cos\theta =cos\alpha \Rightarrow \theta =2n\pi \pm \alpha ]$

$\Rightarrow (x+\frac{\pi }{6})=(2n\pi +\frac{\pi }{3})$ or $(x+\frac{\pi }{6})=(2n\pi -\frac{\pi }{3}).$

$x=(2n\pi +\frac{\pi }{6})$ or $x=(2n\pi -\frac{\pi }{2})$, where $n\in I.$

Hence, the general solution is $x=(2n\pi +\frac{\pi }{6})$ or $x=(2n\pi -\frac{\pi }{6})$, where $n\in I$.