trigonometry
How many solutions has the equation $\sin x= \frac{x}{100}$ ?
Usually when I was asked to solve this type of problem, I would solve it graphically but this one seems to be trickier. It doesn't seem wise to put $f(x)=\sin x$ and $g(x)=\frac{x}{100}$ in the same graph and then counting all the intersection points.
What would be some algebraic methods to solve this?
In the picture below, the red line is all the points where $y=\frac12$. The green line is all the points where $y=\cos x$. Wherever the red and green lines intersect, we have both $y=\frac12$ and $y=\cos x$, so $\cos x = \frac12$.
Since the green line continues wiggling back and forth in the same manner, all the way out to infinity in both directions, and the red line continues straight to infinity in both directions, the red and green lines will intersect an infinite number of times; this shows that the equation $\cos x = \frac12$ has an infinite number of solutions.
Not all trigonometric equations have an infinite number of solutions. For example, $\cos x = 73$ has no solutions. $\cos x = kx$ has a finite number of solutions (except for $k=0$) whose number depends on $k$. For example, $\cos x = \frac1{10}x$ has 7 solutions.
$\sin(2x) = \cos(3x) \Rightarrow \cos(\frac{\pi}{2}-2x) = \cos (3x) \Rightarrow \dfrac{\pi}{2} - 2x = \pm 3x + 2k\pi, k \in \mathbb{Z}$. Can you take it from here?
Best Answer
First, we may suppose $x\ge 0$ since both sides are odd functions.
Using the Intermediate value theorem, there'll be two non-negative solutions on each interval $]2k\pi,2(k+1)\pi[$ as long as $\frac x{100}\le 1$, i.e. $x\le 100$. There results the number of non-negative solutions is equal to $2\times \biggl\lfloor \dfrac{50}{\pi}\biggr\rfloor=32$.
Hence, by symmetry, the total number of roots is $\;\color{red}{63}$.