Expand sin(2x)=2sin(x)cos(x). So 2sin(x)cos(x)=10cos(x). cos(x)(2sin(x)-10)=0. The only solution is cos(x)=0, because sin(x) cannot =5 (sin(x) has a range of -1 to 1). So x=(2n-1)(pi)/2, that is, an odd multiple of (pi)/2 or 90 degrees, where n is an integer. This was a sort of trick question and wasn't as hard as you might have thought. To keep it in the specified range, we have (pi)/2 or 3(pi)/2. That is, 90 and 270 degrees.