sin(x)=⅗, so we have a 3-4-5 Pythagorean Triangle where 5 is the length of the hypotenuse.
cos(x)=⅘. (Pythagoras: 52=32+42.) tan(x)=¾ or tan(x)=sin(x)/cos(x)=⅗/⅘=¾.
Let t=t/1=tan(x/2), tan(x)=2t/(1-t2), sin(x/2)=t/√(1+t2), cos(x/2)=1/√(1+t2) from right triangle with sides 1, t, √(1+t2) (Pythagoras).
tan(x)=2t/(1-t2)⇒¾(1-t2)=2t, 3(1-t2)=8t, 3t2+8t-3=0=(3t-1)(t+3), so t=tan(x/2)=1/3 or -3. We can reject -3, because x would not be in the required range 0-π/2. Therefore, t=tan(x/2)=⅓, sin(x/2)=⅓/√(10/9)=1/√10 which is the same as √10/10. cos(x/2)=1/√(10/9)=3/√10 or 3√10/10 or 0.3√10.