bisection iteration wither different values than 0

How to solv the eq f(y)=0.2? instead of f(y)=0.

 d=0.250; % I SUPPOSED d=1.3
 f=@(y) 0.46-0.5*cos(3.14*(y/d))+0.004*cos(2*3.14*(y/d))
 format long
 eps_abs = 1e-5;
 eps_step = 1e-5;
 a = 0; % Initial Guess to your function such that f(a)>0.
 b = 1; % Initial Guess to your function such that f(b)<0.
 while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) )  >= eps_abs ) )
    c = (a + b)/2;
    if ( f(c) == 0 )
       break;
    elseif ( f(a)*f(c) < 0 )
       b = c;
    else
       a = c;
    end
 end

Answer 1

[Walter Roberson]

To solve f(y) = 0.2, define f1(y) = f(y) - 0.2, and then solve for f1(y) = 0.