You are looking at the equality cos(x)+sin(x)=x2. Let's simplify this a bit and look at cos(x)+sin(x)-x2=0. Take the left hand side of the equation to be f:
We are concerned about what is happening to this function in the interval [0,Π/2]. Well, we know
Notice that f(0) is positive, and f(Π/2) is negavive.
The intermediate value theorem states that if f is a continuous function over the interval [a,b], and if u is a number such that f(a)>u>f(b), then there exists some c∈[a,b] such that f(c)=u.
Taking a step back and putting this to use, we are using f(x)=cos(x)+sin(x)-x2 over the interval [0,Π/2], and we are saying that 0 is a number such that
Hence, by the intermediate value theorem, we know that there exists some real number c∈[0,Π/2] such that f(c)=0.
But if f(c)=0, then
So the statement is true, there exists at least one point c between 0 and Π/2 such that cos(c)+sin(c)=c2, by the intermediate value theorem.