How does having a repeated root in $trigonometric$ $equations$ (non-polynomial) imply touching, but not crossing the $x$-axis?

Question

Consider

$$f(x)=cos\left(2x+\frac{\pi}{3}\right)-cos\left(\frac{3x}{2}+\frac{\pi}{4}\right)=0 \tag{1}$$ giving $$2x+\frac{\pi}{3}=\pm\left(\frac{3x}{2}+\frac{\pi}{4}\right)+2r\pi \tag{2}$$ $\pm \to +, r \to n$ $$x_n^+=\pi\left(\frac{24n-1}{6}\right) \tag{3}$$ $\pm \to -, r \to k$ $$x^-_k=\frac{\pi}{42}\left(24k-7\right) \tag{4}$$ where $r,n,k \subset Z$.

Then $x=...,-31\pi/42,-\pi/6,17\pi/42, 41\pi/42...,$

According to the textbook that I am reading, as $$x^+_{n=0}=x^-_{k=0}=-\frac{\pi}{6} \tag{5}$$ is a repeated root, $x=-\frac{\pi}{6}$ is a value at which the curve touches the $x-axis$ without crossing it.

But I do not understand how having a repeated root in trigonometric equations implies touching the $x$-axis (the notion of change of sign does not seem applicable here)?

Answer 1

Any real function $y=f(x) $ whether algebraic or trigonometric, has a double root at $x=a$ if

$$ f(a)=0,\; f'(a)=0\;$$

For example there is a double root at

$x=-1$ for the curve (circle)

$$ (x+1)^2+(y+1)^2=1$$

where it touches x-axis. A trig function also has the same situation:

Quite the same happens for the following implicit trig function

$$ \sin x \sin (y-\frac14) = \frac14 $$

Any smooth function can be chosen from trigonometric, hyperbolic algebraic polynomials functions to be eligible.They should have series expansion with continuous derivatives. You were perhaps first introduced to algebraic polynomials for ease of differentiation.