Pre calculus question

Question

We can find the solutions of sin x = 0.5 algebraically. (Round your answers to two decimal places.)

(a) First we find the solutions in the interval [0, 2 pi). We get one such solution by taking sin^-1 to get x= ? (Smaller value). The other solution in this interval is x= ? (Larger value).

(b) We find all solutions by adding multiples of ? to the solutions in [0, 2 pi). The solutions are x = ? (Enter your answers in the form thetha + 2pi k, 0 ≤ 0 < 2pi. Enter your answers as a comma-separated list.)

Answer 1

We can find the solutions of sin x = 0.5 algebraically. (Round your answers to two decimal places.)

(a) First, we find the solutions in the interval [0, 2 pi).  We get one such solution by taking sin^-1 (on a calculator) to get x = 0.5236 radians (0.52 rounded) = π/6 radians (or 30º). Note that the inverse sine function is evaluated from -π/2 to π/2 only – the first & fourth quadrant.

The other solution in this interval [0,2π) is in the second quadrant where the sine is also positive.  Since the reference angle is π/6, the second quadrant angle is therefore: π ‑ π/6 = 5π/6 and x= 2.6180 radians (2.62 rounded) = 5π/6 radians (or 150º).

(b) We find all solutions by adding multiples of 2π radians (360º) to the solutions in [0, 2 pi). The solutions are x = π/6 + 2πk, 5π/6 + 2πk (x=30º + 360ºk, 150º + 360º). (you can substitute the rounded values if you wish.)