Well hey I need a hand solving the equation 2sinx+1=b, where b is positive real number and has 1 solution between [0,2pi], so whats the value of b?
Sorry to annoy u guys, I couldn't find a unit 1/2 thread 
If it's going to have 1 solution between [0, 2pi] then your 'x'-values' will either be pi/2 or 3pi/2 which you will sub in to find when 'b' is a positive real number (that is b=3 which is your final answer).
I guess to picture this is when you have a unit circle and you go around the circle at x=0, pi/2, pi , 3pi/2 and 2pi, what do they equal when you substitute the 'x-value' in sin(x)? You'll find that at x=0 and x=2pi equal the same thing (hence eliminating those 'x-values as they both equal zero whilst the others give you values of 1 and -1).
Using your exact values won't work since you only want 1 solution between [0, 2pi] as your exact values will give you 2 solutions.. like sin(pi/3) is the same as sin(2pi/3) yet they equal the same value so it'll give you two solutions leaving you with only -1 and 1 as values which you can sub in... hope that makes sense!