Question

The general solution for the equation $\sqrt{3}\sin x-\cos x=2$ is

• A. $(6n+1)\frac{\pi }{6}+(-1{)}^n\frac{\pi }{2},n\in Z$
• B. $(6n-1)\frac{\pi }{4}+(-1{)}^n\frac{\pi }{2},n\in Z$
• C. $(6n+1)\frac{\pi }{4}+(-1{)}^n\frac{\pi }{2},n\in Z$
• D. $(6n-1)\frac{\pi }{6}+(-1{)}^n\frac{\pi }{2},n\in Z$

Answer 1

The correct option is A $(6n+1)\frac{\pi }{6}+(-1{)}^n\frac{\pi }{2},n\in Z$

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

Comparing with the standard equation $a\sin x-b\cos x=c,$

$a=\sqrt{3}\ne 0;b=3\ne 0;c=2;$
$\frac{c}{\sqrt{a^2+b^2}}=\frac{2}{\sqrt{(\sqrt{3}{)}^2+1^2}}=\frac{2}{\sqrt{4}}=1$

$\therefore -1\le \frac{c}{\sqrt{a^2+b^2}}\le 1,$ which signifies the given eqaution has a valid solution.

Divide the equation by $\sqrt{a^2+b^2}=2.$
$\Rightarrow \frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x=\frac{2}{2}$
Here, the auxiliary angle is $\frac{\pi }{6}.$

Substitute $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\sin \frac{\pi }{6}=\frac{1}{2}$
$\Rightarrow \cos \frac{\pi }{6}\sin x-\sin \frac{\pi }{6}\cos x=1$
Apply $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\Rightarrow \sin (x-\frac{\pi }{6})=1=\sin \frac{\pi }{2}$

The general solution is $x-\frac{\pi }{6}=n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$
$\Rightarrow x=n\pi +\frac{\pi }{6}+(-1{)}^n\frac{\pi }{2},n\in Z$
$\Rightarrow x=(6n+1)\frac{\pi }{6}+(-1{)}^n\frac{\pi }{2},n\in Z$