Show $2\sin(x) + \cos(x) = x$ has three solutions. Using IVT.
Attempt:
After drawing the function on a graph I could observe that there are three solutions. So my first idea was to use the same idea used in showing that there exists $x \in (0,\frac{\pi}{2})$ such that $\cos(x) = x$.
Applying this idea over the interval $[0,\pi]$ and defining $f(x) = 2\sin(x) + \cos(x) - x$, we obtain $f(0) = 3$ and $f(\pi) = -(3 + \pi) < 0$. So by IVT there exists an $x \in [0,\pi]$ such that $2\sin(x) + \cos(x) = x$.
Comment:
So I tried to extend this to the negative side, but the same idea failed. I have two questions:
1) Is this the right approach to take and maybe my arithmetic is off?
2) I arbitrarily picked the interval $[0, \pi]$ because I had a feeling based on the trig identities, but how would I go about selecting intervals if I wasn't aware of this? There must be a more efficient way.
The Intermediate Value Theorem can only show that you have at least three solutions. To show that there can be no more requires further work.
So, you were right to start with $f(x)$. I'm not sure what you did with your arithmetic or trig values. Let's make a little table: \begin{matrix} x & f(x) \\ \hline -\pi & -1+\pi > 0 \\ -\pi/2 & -2+\pi/2 < 0 \\ 0 & 1 > 0 \\ \pi & -1-\pi < 0 \end{matrix} (I think your values for $0$ and $\pi$ were wrong. So it makes me suspect any computation you were doing! :P) This shows three sign changes; hence, since $f$ is continuous, the IVT guarantees at least three zeroes.