What are the polar coordinates of $(2\sqrt3, 2)$?

Question

My answer to this is $(4,\frac{π}6)$. But a calculator said that $(-4,\frac{7π}6)$ is also an answer, and there are infinitely many solutions. Is that correct?

Answer 1

Your solution is correct. Now, concerning $\left(-4,\frac{7\pi}6\right)$, it is indeed true that$$-4\left(\cos\left(\frac{7\pi}6\right),\sin\left(\frac{7\pi}6\right)\right)=\left(2\sqrt3,2\right),$$but $-4<0$, and therefore that answer is wrong. And your answer is the only one which uses an angle from $[0,2\pi)$.