functionstrigonometry
Hello everyone how can I find the count of the solution for $\sin(x) = x^2$?
I know there is a one solution in $x = 0$ and for the other solutions I tried to find the extreme point of the function: $y = x^2 – \sin(x)$ and $y'$ is:
$y' = 2x -\cos(x)$ but I don't know how to solve this equation.
A graph indicates eight solutions. Four are $\pm \frac {\pi}2$ and $\pm \frac {3\pi}2$ where the value of both expressions is $1$. You can then observe that at $-2\pi,0,2\pi, 3^{\cos (x)} = 3 \gt |\sin (x)|$, and at $-\pi, \pi,\ 3^{\cos (x)} \gt |\sin (x)|=0$
Solution to $\cos \alpha = 0$ is $$\alpha = {\pi \over 2} +\pi n$$
so in your case $$x ={\pi \over 8} +{\pi \over 4}n$$
$n\in\mathbb{Z}$.
Best Answer
It's quite obvious that there are no solutions when $x<0$, so we will look for $x\ge0$. You have found that $x=0$ satisfies the equation. Let's analyze for $x>0$:
Take $f(x)=x^2$ and $g(x)=\sin(x)$.
For $x=\frac{\pi}{4}$, some calculations give $f(\frac{\pi}{4})\approx 0.625$ while $g(x) \approx 0.7$: $$f(\frac{\pi}{4}) < g(\frac{\pi}{4})$$
For $x=1$, $f(1)=1$ but $g(1)<1$ since $\sin(x)$ is increasing for $x\in[0,\pi/2]$ and $\sin(\pi/2)=1$, then $$f(1)>g(1)$$ which means $f(x)$ exceeds $g(x)$ between $(\pi/4,1)$ and intersect in this inteval. Now you just need to prove that they can't intersect more than once.