The question is show that the equation f(x)=3x-2+cos(Piex/2)=0 has exactly one real root. Using the intermediate value theorem f is continuous and f(o)=-1 and f(1)=1., the equation has one root. Supposing it has two roots a and b , and a<b. Then f(a)=0=f(b). Using Rolle's theorem there is a number d in (a,b) such that f'(x)=3-(pie/2)(sin(piex/2))=0 has a root in (a,b). But this is impossible since after factoring sin(piex/2)=6/pie is always greater than zero for all real x. I am having problems with factoring. Why is 3-(pie/2)(sin(piex/2))always >o for all x?