Determine the number of solution in the interval 0≤x≤2π for sin(ax)=0.5, where a is an integer and a≥1

Note that, from the unit circle, sin(x) = 1/2 when x = π/6 ± 2πk and x = 5π/6 ± 2πk where k is an integer.

Thus:
sin(ax) = 1/2
==> ax = π/6 ± 2πk and ax = 5π/6 ± 2πk
==> ax = π(12k ± 1)/6 and ax = π(12k ± 5)/6, by combining fractions
==> x = π(12k ± 1)/(6a) and x = π(12k ± 5)/(6a), by dividing both sides by x.

Since we only want solutions on 0 ≤ x ≤ 2π, k ≥ 0 (as k < 0 produces negative values of x, which lie outside of the range). Thus, we can drop the plus-or-minus sign and replace it with a plus sign to give:
x = π(12k + 1)/(6a) and x = π(12k + 5)/(6a)

For the first solution set, we see that:
0 ≤ x ≤ 2π ==> 0 ≤ π(12k + 1)/(6a) ≤ 2π.

Multiplying all parts by 6a yields:
0 ≤ π(12k + 1) ≤ 12πa.

Dividing all parts sides by π:
0 ≤ 12k + 1 ≤ 12a ==> 0 ≤ 12k ≤ 12a - 1.

Dividing all parts by 12:
0 ≤ k ≤ (12a - 1)/12 ==> 0 ≤ k ≤ a - 1/12.

Since k is an integer k is any integer between 0 and a - 1 (a solutions). Thus, there are (a - 1) + 1 = a solutions to this part.

In a similar fashion, x = π(12k + 5)/(6a) lies on 0 ≤ x ≤ 2π if:
0 ≤ k ≤ (12a - 5)/12 ==> 0 ≤ k ≤ a - 5/12,

which also has a integral solutions.

Therefore, there are a + a = 2a solutions as required.

I hope this helps!

"Since k is an integer k is any integer between 0 and a - 1 (a solutions). Thus, there are (a - 1) + 1 = a solutions to this part."
I don't get it from here. could u please explain a little bit more?

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0 ≤ x ≤ 2π so 0 ≤ ax ≤2aπ Noe sinax = 0.5 means ax = π/6 + 2nπ or (2n+1)π - π/6, in general.
so x = 2nπ/a + π/6a or (2n+1)π/a - π/6a
So a must be a divisor of n and there will me many solutions.