An algebraic argument: The fact that a polynomial of degree $n$, where $n \ge 1$, has at most $n$ roots can be proved without using machinery from the calculus. When $n=1$, the result is clear, there is in fact precisely one root. Suppose now that we know that the result is true for any polynomial of degree $k$. We show that the result is true for any polynomial of degree $k+1$.
Let $P(x)$ have degree $k+1$. If $P(x)$ has no roots, we are finished. If $P(x)$ has a root $\alpha$, then the polynomial $x-\alpha$ divides $P(x)$. So $P(x)$ is identically equal to $(x-\alpha)Q(x)$ for some quotient polynomial $Q(x)$. Now $Q(x)$ has degree $k$, so by assumption has no more than $k$ roots. It follows that $P(x)$ has at most $k+1$ roots, namely the roots of $Q(x)$, together with $\alpha$.
The above proof works for polynomials with coefficients in any field.
A calculus argument: For polynomials with real coefficients, we can prove the result by using "calculus" tools. These are not really appropriate, since the calculus tools are far more sophisticated than the algebraic tools used in the above proof, and we end up with a proof that only applies to polynomials with real coefficients. But as an exercise, let's do it.
The tool that gives the quickest argument is the Mean Value Theorem, actually a special case of it usually called Rolle's Theorem. This says that if $f(x)$ is a function that is differentiable in the interval $[a,b]$, and $f(a)=f(b)=0$, there is a $c$, with $a<c<b$, such that $f'(c)=0$. Informally, between any two roots of $f(x)$, there is a root of the derivative $f'(x)$.
Back to polynomials. Suppose that we know that any polynomial of degree $k$, with real coefficients, has at most $k$ real roots. We want to show that a polynomial $P(x)$ of degree $k+1$ has at most $k+1$ real roots.
Suppose to the contrary that $P(x)$ has at least $k+2$ roots. Pick any $k+2$ of them, and list them in increasing order $\alpha_1$, $\alpha_2$, and so on up to $\alpha_{k+2}$. By Rolle's Theorem, between any two roots of $P(x)$, there is a root of $P'(x)$. That implies that $P'(x)$ has at least $k+1$ roots. But $P'(x)$ is a polynomial of degree $k$, so by assumption can have no more than $k$ roots. This completes the proof.
We can avoid Rolle's Theorem, and use only the Intermediate Value Theorem. We want to show that between any two roots of $P(x)$, there is a root of $P'(x)$. The easiest calculation uses the fact that if $\alpha$ is a root of $P(x)$, then $x-\alpha$ divides $P(x)$. It is slightly unpleasant, since we have to take special care with roots of multiplicity greater than $1$.
The Fundamental Theorem of Algebra: I would guess that what you were told is a distorted version of an important comment about proofs of what is usually called the Fundamental Theorem of Algebra. Roughly speaking, this result says that a polynomial of degree $n \ge 1$ with complex coefficients has, if you take multiplicities into account, exactly $n$ roots in the field $\mathbb{C}$ of complex numbers.
There is a large number of different proofs of this result. The classical ones depend on fairly subtle arguments about functions of two variables.
The end of the Wikipedia article has a list of useful references.
The following important result is a consequence of the Fundamental Theorem of Algebra. Conversely, the Fundamental Theorem of Algebra can derived fairly simply from it. The result mentions only real numbers.
Theorem: Let $P(x)$ be a polynomial of degree $\ge 1$, with real coefficients. Then $P(x)$ can be expressed as a product $A_1(x)A_2(x)\cdots A_k(x)$, where each $A_i(x)$ has real coefficients, and is of degree $1$ or $2$.
There are proofs of this that use a fair bit of algebra (in the modern sense of algebra) and very little function theory. The only bit of "calculus" needed is to to show that any odd degree polynomial has a real root, and any positive number has a square root, both of which are consequences of the Intermediate Value Theorem. Proofs that use so "little" machinery from analysis tend to boast about it by saying that they "only" use the Intermediate Value Theorem.
You can't use IVT to verify that those points are solutions. To do that, simply plug them in and observe that they satisfy the equation.
The way IVT is used is to prove that a solution must exist. This is very different than directly finding a solution, as you have done. To use IVT in this problem, first move everything to one side of the equation so that we have $$f(x) = e^{\sin(x)} - 2 - \cos(x) + \sin(x)$$ Now plug in the values $x=\pi/2, 3\pi/2$ and observe that $f(\pi/2) = e - 2 + 1 > 0$, while $f(3\pi/2) = e^{-1} - 2 - 1 < 0$. Therefore, by the IVT, there exists a number $c \in (\pi/2, 3\pi/2)$ such that $f(c) = 0$. The IVT doesn't give us any idea what $c$ is; it just tells us that $c$ exists.
Actually, your solution is much better than using IVT, because it is obviously more useful to know what some of the solutions are, rather than simply knowing that the solutions exist.
Best Answer
Assuming $x\neq 1$, the equation you have to solve is $$x^5-11 x^4+1000=0$$ So, consider the function and derivatives $$f(x)=x^5-11 x^4+1000$$ $$f'(x)=5x^4-44 x^3$$ $$f''(x)=20x^3-132x^2$$ The first derivative cancels at $x=0$ and $x=\frac{44}5$. $$f(\frac{44}5)=-\frac{38104056}{3125}<0$$ and the second derivative test shows that this is a minimum; so you have two real roots.
If you deflate the quintic polynomial, you are let with a cubic which has at least one real root.
So, you have at least three real roots that you properly located.