(a) f(c)=V=4 then c²-c=4. Let c=2, f(2)=4-2=2, f(5)=25-5=20, therefore, since 2<4<20 (4 is between 2 and 20), we know that c exists in the interval. f(3)=9-3=6, so c is between 2 and 3. f(2.5)=6.25-2.5=3.75 which is less than V. So c is between 2.5 and 3. c²-c-4=0, c=(1±√(1+16))/2. But we need the value between 2.5 and 3, so c=(1+√17)/2=2.56155... f(½(1+√17))=4=V. The other value for c is not in the given interval.
(b) c=π/6, cos(0)=1 and cos(π/2)=0, π/6 is between 0 and π/2.