Question

Find the angles between -pi and pi that satisfy:

(−√3) sin(v)+cos(v)=√3

Answer 1

Observe that

`\sin\left(\frac\pi6-v\right)=\sin\frac\pi6\cos v-\cos\frac\pi6\sin v=\frac12\cos v-\frac{\sqrt3}2\sin v`

In the original equation, divide both sides by
`\frac12`:

`-\sqrt3\sin v+\cos v=\sqrt3\implies\frac12\cos v-\frac{\sqrt3}2\sin v=\frac{\sqrt3}2`

`\implies\sin\left(\frac\pi6-v\right)=\frac{\sqrt3}2`

Next,

`\sin x=\frac{\sqrt3}2\implies x=\frac\pi3+2n\pi,x=\frac{2\pi}3+2n\pi`

where
`n` is any integer. Then

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

Fix
`n=0` to ensure
`-\pi<v<\pi`, so that

`v=-\frac\pi6,v=-\frac\pi2`