Angel P. answered 06/01/23
MIT Student Tutor
In order to solve this problem, we should recall our conversion factors for polar and rectangular coordinates: x = rcos(θ) and y = rsin(θ). Using our given values, we know x = -2sqrt(3) = rcos(θ) and y = 2 = rsin(θ). We can then apply our knowledge of trigonometric identities, namely tan(θ) = sin(θ) / cos(θ). We can divide our two equations to find y / x = rsin(θ) / rcos(θ) = tan(θ) = 2/(-2sqrt(3)) = -1/sqrt(3). This leaves us with tan(θ) = -1/sqrt(3). We can then solve for θ by taking the inverse tangent of both sides, giving θ = arctan(-1/sqrt(3)) = -π/6. We now know that θ = -π/6. We can use this information to then solve for r by plugging θ back into either of our original conversions. We can then take y = 2 = rsin(-π/6), divide both sides by sin(-π/6), and arrive at r = 2/sin(-π/6) = -4. We can then arrive at our polar coordinates, (r, θ) = (-4, -π/6). However, we would like our r value to be positive and so we can multiply our r by -1 and flip our angle 180 degrees by adding π. Therefore, we arrive at (r, θ) = (4, 5π/6) = (-4, -π/6).