Since no one else seems to be taking the analytic approach, I'll present the solution I recall.

sin^10(x) + 5*sin^8(x)*cos^2(x) + 10*sin^6(x)*cos^4(x) + 10*sin^4(x)*cos^6(x) + 5*sin^2(x)*cos^8(x) +cos^10(x) = 1

Now take the difference of this identity and the equation we want to solve, sin^10(x) + cos^10(x) = 61/256, and simplify:
sin^8(x)*cos^2(x) + 2*sin^6(x)*cos^4(x) + 2*sin^4(x)*cos^6(x) + sin^2(x)*cos^8(x) = 39/256.

Now we'll factor:

sin^2(x)*cos^2(x) * [sin^6(x) + cos^6(x) + 2*sin^4(x)*cos^2(x) + 2*sin^2(x)*cos^4(x)] = 39/256

Then to deal with sin^6(x) + cos^6(x), we will take the third power of sin^2(x) + cos^2(x) = 1, and rearrange:

sin^6(x) + cos^6(x) = 1 - 3*sin^4(x)*cos^2(x) - 3*sin^2(x)*cos^4(x)

Then substitute this in and factor one more term out:

sin^2(x)*cos^2(x) * [1 - (sin^2(x)*cos^2(x)) * (sin^2(x) + cos^2(x))] = 39/256.

One last substitution of sin^2(x) + cos^2(x) = 1 and then we can also substitute sin(x)*cos(x) = (1/2)*sin(2x).  Plugging those in the equation can be reduced to:

(1/4)*sin^2(2x) - (1/16)*sin^4(2x) = 39/256.

This is a quadratic equation in sin^2(2x).  Then for clarity if y=sin^2(2x) then our quadratic is (1/16)y^2 - (1/4)*y + 39/256 = 0.

This has two roots, y=3/4 and y=13/4.  13/4 must be discarded since the range of sin^2(2x) is 0 to 1.  Then sin^2(2x) = 3/4.

Then sin(2x) = sqrt(3)/2 or -sqrt(3)/2, which makes 2x=pi/3 + 2k*pi, or 2x=2pi/3 + 2k*pi, or 2x=4pi/3 + 2k*pi, or 2x=5pi/3 + 2k*pi, where k is an integer.

2x=pi/3 + 2k*pi and 2x=4pi/3 + 2k*pi are exactly half a period offset, so they can be combined into 2x = pi/3 + k*pi.  Similarly, 2x=2pi/3 + 2k*pi and 2x=5pi/3 + 2k*pi can be combined into 2x = 2pi/3 + k*pi.

Dividing through by 2, we get the final answer of x=pi/6 + (k/2)*pi or x=pi/3 + (k/2)*pi, where k is an integer