how many solutions are there?

Question

Consider the following family of equations:
sin(nx) = sin((n + 1)x), n = 1, 2, 3, . . . .
From the case when n = 1, 2, . . . , 10, can you guess how many
solutions are there to the equation
sin(1101x) = sin(1102x)
on the interval [−pi, pi]?

Answer 1

sin(x) = sin(2x) has 5 solutions
sin(2x) = sin(3x) has 7 solutions
sin(3x) = sin(4x) has 9 solutions
sin(4x) = sin(5x) has 11 solutions
..........
There seems to be a pattern here, where
sin(nx) = sin[(n + 1)x] has (2n + 3) solutions.

If n = 1101, then I guess there are
2*1101 + 3 = 2,205 solutions.

Answer 2

Let's start with n = 1.
Then sin x = sin 2x = 2 sin x cos x.
So if sin x = 0
x = -π, 0, π.
If sin x <> 0 then 2 cos x = 1
cos x = 1/2
x = π/3 or -π/3.
So there are 5 solutions
Now let's investigate n = 2.
sin 3x = sin 2x cos x + cos 2x sin x
= 2 sin x cos²x + cos 2x sin x = sin 2x = 2 sin x cos x.
Again, sin x = 0 yields x = -π, 0 , π.
Else divide out sin x and simplify to get
2 cos² x + 2 cos² x -1 = cos x
or
4 cos² x - cos x -1 = 0.
so
cos x = 1/8[ 1 + √17].
or
cos x = 1/8[ 1 - √17].
So there are 7 solutions in this case.
Update: I worked out the case sin 3x = sin 4x.
It leads to the equation
8 cos³ x - 4 cos² x -4 cos x +1 = 0.
By using approximate solving methods or Sturm's theorem
we find that this equation has 3 real roots in cos x,
so there are 9 solutions of the equation
in [-π, π].
So it seems that the pattern yields 2n + 3 solutions
for general n and that for n = 1101 we get 2205
solutions in [-π, π].
I will investigate the remaining cases later, but
I don't see how to prove the pattern in general.
If we try to solve the general case it would
be nice to know:
a). Are all the coefficients of the resulting
polynomial in cos x powers of 2?
b) Are all the roots of this polynomial real?
An interesting problem!