Question

Find the angles between -pi and pi that satisfy: (−√3) sin(v)+cos(v)=√3

Answer 1

Observe that

$\sin\left(\dfrac\pi6-v\right)=\sin\dfrac\pi6\cos v-\cos\dfrac\pi6\sin v=\dfrac12\cos v-\dfrac{\sqrt3}2\sin v$

In the original equation, divide both sides by $\dfrac12$:

$-\sqrt3\sin v+\cos v=\sqrt3\implies\dfrac12\cos v-\dfrac{\sqrt3}2\sin v=\dfrac{\sqrt3}2$

$\implies\sin\left(\dfrac\pi6-v\right)=\dfrac{\sqrt3}2$

Next,

$\sin x=\dfrac{\sqrt3}2\implies x=\dfrac\pi3+2n\pi,x=\dfrac{2\pi}3+2n\pi$

where $n$ is any integer. Then

$\sin\left(\dfrac\pi6-v\right)\implies v=-\dfrac\pi6-2n\pi,v=-\dfrac\pi2-2n\pi$

Fix $n=0$ to ensure $-\pi$, so that

$v=-\dfrac\pi6,v=-\dfrac\pi2$