Observe that
[tex]\sin\left(\dfrac\pi6-v\right)=\sin\dfrac\pi6\cos v-\cos\dfrac\pi6\sin v=\dfrac12\cos v-\dfrac{\sqrt3}2\sin v[/tex]
In the original equation, divide both sides by [tex]\dfrac12[/tex]:
[tex]-\sqrt3\sin v+\cos v=\sqrt3\implies\dfrac12\cos v-\dfrac{\sqrt3}2\sin v=\dfrac{\sqrt3}2[/tex]
[tex]\implies\sin\left(\dfrac\pi6-v\right)=\dfrac{\sqrt3}2[/tex]
Next,
[tex]\sin x=\dfrac{\sqrt3}2\implies x=\dfrac\pi3+2n\pi,x=\dfrac{2\pi}3+2n\pi[/tex]
where [tex]n[/tex] is any integer. Then
[tex]\sin\left(\dfrac\pi6-v\right)\implies v=-\dfrac\pi6-2n\pi,v=-\dfrac\pi2-2n\pi[/tex]
Fix [tex]n=0[/tex] to ensure [tex]-\pi<v<\pi[/tex], so that
[tex]v=-\dfrac\pi6,v=-\dfrac\pi2[/tex]
