The obvious thing seems to be to expand sin(7x). Its possible to do all that by hand using repeated application of angle-sum formulas, or use Chebyshev polynomials to short-cut your way there.
Either way sin(7x) = -64(sin x)^7 + 112(sin x)^5 - 56(sin x)^3 + 7*sin(x).
Then we have -64(sin x)^7 + 112(sin x)^5 - 56(sin x)^3 + 7*sin(x) = 7*sin(x).
Which simplifies to (sin x)^3 * [8(sin x)^4 - 14(sin x)^2 + 7] = 0.
So one set of solutions is given by sin x=0, or x=pi*n for integer n.
The quartic polynomial 8z^4-14z^2+7 has four complex roots: r = +/-[(7+/-sqrt[7])/8]
So if we assume x is real then (sin x)^2 needs to be a real number in [0,1]; so there are no new real solutions.
Otherwise in complex numbers we have four additional sets given from x=arcsin(+/-[(7+/-sqrt[7])/8]) + pi*k, for each possible choice of +/-.
Solution
