Find the roots of the equation sin (2x-n / 2) = – 1/2 belonging to the half-interval (0; 3n / 2]

Let us find the roots of the equation sin (2 * x – n / 2) = – 1/2 belonging to the half-interval (0; 3 * n / 2].

sin (2 * x – n / 2) = – 1/2;

2 * x – pi / 2 = (- 1) ^ n * arcsin (- 1/2) + pi * n, where n belongs to Z;

2 * x – pi / 2 = (- 1) ^ n * 7 * pi / 6 + pi * n, where n belongs to Z;

We transfer the known values ​​to one side, and the unknown ones to the other side. When transferring values, their signs change to the opposite sign. That is, we get:

2 * x = (- 1) ^ n * 7 * pi / 6 + pi / 2 + pi * n, where n belongs to Z;

x = (- 1) ^ n * 7 * pi / 12 + pi / 4 + pi * n / 2, where n belongs to Z;

For n = 0, x = 7 * pi / 12 + pi / 4 = 5 * pi / 6 belongs to (0; 3 * n / 2];

For n = 1, x = – 7 * pi / 12 + pi / 4 + pi / 2 = 2 * pi / 3 belongs to (0; 3 * n / 2];

For n = 2, x = 7 * pi / 12 + pi / 4 + pi = 13 * pi / 3 does not belong to (0; 3 * n / 2];

For n = – 1, x = – 7 * pi / 12 + pi / 4 – pi / 2 = – 5 * pi / 3 does not belong to (0; 3 * n / 2];

Answer: x = 5 * pi / 6 and x = 2 * pi / 3.