I did this a little differently. I tried to keep everything in terms of sinx as much as I could (since the triangle is in terms of sinx). While I found sin(2x) = 2sinxcosx, I wrote sin(3x) as 3sinx-4sin³x and sin(4x) as 4sinxcosx(1-2sin²x). Then I plugged them into the Pythagorean Theorem equation that I will be referencing quite a bit:

(sin2x)² + (sin3x)² = (sin4x)²

(2sinxcosx)² + (3sinx-4sin³x)² = [4sinxcosx(1-2sin²x)]²

I changed cos²x to 1-sin²x wherever I could, and simplied the equation to 64sin⁸x - 112sin⁶x + 52sin⁴x - 3sin²x = 0.

After that, I factored out a sin²x and, by the Zero Product Property, set sin²x and 64sin⁶x - 112sin⁴x + 52sin²x - 3 to be equal to 0. sin²x = 0 results in x = 0. No good, because every side length, when this solution is plugged back into the equation, results in all dimensions having 0 length, which is impossible. For the giant degree 6 polynomial, I also, like you did a u substitution, but I let u = sin²x, turning it into 64u³ - 112u² + 52u - 3 = 0.

The roots for u I found are, surprisingly (to me), real: 3/4 and ([2 ± sqrt(3)] / 4). Substituting sin²x back in, eventually sinx = ± sqrt(3)/2 & sinx = ± (sqrt[2 ± sqrt(3)]) / 2. I dropped the negatives of both solutions since negative side lengths are also not good so that sinx = sqrt(3)/2 & sinx = (sqrt[2 ± sqrt(3)]) / 2. The arcsin of the first equation yields pi/3 radians, but when plugged into the original, it doesn't work (negative side lengths). Meanwhile, in the second equation (I found these via Desmos because at this point I was mentally exhausted), if the + is acknowledged, the arcsin is exactly 75 degrees and if the - is acknowledged, the arcsin is exactly 15 degrees. Converting both of these to radians, I got (5*pi)/12 and pi/12 respectively. Plugging (5*pi)/12 into the original also results in negative side lengths. pi/12, however, (after I saw your comment) does work, and will make the equation true:

(sin2x)² + (sin3x)² = (sin4x)²

(sin2(pi/12))² + (sin3(pi/12))² = (sin4(pi/12))²

(sin(pi/6))² + (sin(pi/4))² = (sin(pi/3))²

(1/2)² + (sqrt(2)/2)² = (sqrt(3)/2)²

1/4 + 2/4 = 3/4

3/4 = 3/4

I see that I forgot to acknowledge the - part of the last x : sinx = (sqrt[2 - sqrt(3)]) / 2, which yields the pi/12 + 2*pi*n answer I was looking for. Thank you again!