f(x) = sin x. The equation f(x) = k has two solutions in [0, 2π]. One of the solutions is x = a.

Question

1. What is the other solution?
2. How many solutions would f(bx) = k have in [0, 2π]?

Answer 1

1.for a in (0, π), the other solution is π - a; for a in (π, 2π), the other solution is 3π - a.

We have two solutions for k in (-1,0) ∪ (0, 1); For k = 0, there are 3 solutions, (0, π, 2π)

for k = 1, there is only 1 solution, π/2, and for k = -1, there is only 1 solution, 3π/2.

Take the unit circle and cut it with a horizontal line and see where the circle is cut. Then, consider the angles from the origin where the cuts occur. k = 0 is tricky, because the region was defined as [0, 2π], and 0 and 2π occur at the same point (x should probably have been in [0, 2π)

2.

For k in (-1,0) ∪ (0, 1), if b is an integer, there will be 2|b| solutions. For k = 0, we will have 2|b| + 1 solutions.

For k in (-1,1), we will have |b| solutions.