In which point the tangent of $y= \sin x - \cos x$ is parallel with $y = x$?

Question

Isn't the graph of $y = \sin x - \cos x$ a continous "set of waves". That means that there are several points where the tangent of the function is parallel with $y = x$. That's how I understand it at least. Could someone explain, how exactly can you find this? Thank you

Answer 1

You need to solve $$(\sin{x}-\cos{x})'=1,$$ which is $$\cos{x}+\sin{x}=1$$ or $$\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt2}...$$

Answer 2

the slope of tangent of a function can be calculated with its derivative$$y_1'=[\sin(x)-\cos(x)]'=\cos(x)+\sin(x)\\y_2'=x'=1$$ now just compare them: $$\cos(x)+\sin(x)=1$$you will get $x=\begin{cases}2\pi n\\2\pi n+{\pi\over2}\end{cases},\,n\in\Bbb Z$