Can k - [ ( sin 2k ) / 2] = 5pi / 12 - 1/4 be solved?

[grapz]

The only method i can think of is by guessing the answer to this equation.

k - [ ( sin 2k ) / 2] = 5pi / 12 - 1/4

I guess the answer to be 5pi / 12 and it works

 

[tkhunny]

The 'k' term keeps it from having more than one solution.

Various iterative techniques can be used to find a numeric solution.