When solving a trigonometric equation, like sin(x) = -1/3 for 0 ≤ x < 2π, why do I get an answer outside the range? Why are there many correct answers for the value of x?

Drawing the graphs of the trigonometric functions sin, cos & tan, we can see that they all extend along the x-axis to infinity. Indeed, the sin & cos functions move between 1 and -1 on the y-axis, and extend infinitely along the x-axis. Considering the function y = sin(x) for a given value of y, like y = -1/3, there are infinite solutions of x, as y is equal to -1/3 at multiple points along the function. That is why exam questions limit the range of the x-values they are looking for. Now, solving x = sin^-1(-1/3) gives us x = - 0.3398369095..., which is outside the range 0 ≤ x < 2π, as that is the closest to the origin where sin(x) takes the value -1/3. However, there are solutions for this function within our range. In the graph of y = sin(x), we see that y becomes negative between π and 2π. The value of x at the first time y = -1/3 in our range is π minus the x value which we found before, so x = π - ( -0.3398369095...) = 3.48 radians correct to 2 d.p. The value of x the second time y = -1/3 is in our range is 2π plus the x value we found before, so x = 2π - 0.3398369095... = 5.94 rad to 2 d.p.