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Now I want to apologize in advance for not knowing how to use mathematical expressions here and would appreciate if someone is willing to edit this.

A point is moving on a circle with radius of 2 meters and angular velocity of $\frac12 \text{ rad}\cdot s^{-1}$, with starting angle of $\frac{\pi}{4}$.

In which moment $t \in [0;\frac{\pi}{2}]$ cosine of angle between tangent on $(x(t),y(t))$ and line $y= \Biggl( \frac{-1}{3^{1/2}}\Biggr) x + 7$ will be biggest (maximum).

Now coordinates for the point can be found with formulas and we get $$x(t)=2\cos \biggl(\bigl(\frac12\bigr)t + \frac\pi4\biggr)$$ and $$y(t)=2\sin\biggl(\bigl(\frac12\bigr)t + \frac \pi4\biggr).$$

I went and found the derivate of both $x(t)$ and $y(t)$ and we find the tangent $$y'(t)= \cot \biggl(\bigl(\frac12\bigr)t + \frac \pi4\biggr).$$ Now I'm stuck and I don't really know how to set up the function in which we can find maximum of and angle. Thank you in advance for any kind of answer or hint.

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First of all, some corrections. The derivative that you found in the end of your post is NOT $y'(t)$ — it's $y'(x)=\frac{dy}{dx}$. And it's slightly wrong — a negative sign is missing. Do you see why?

Now, what is the maximal possible value of cosine? And what value of an angle, $\theta=\ldots$, has this maximal possible value of cosine? That's the angle you want between the two lines (the given line and the tangent line). Then think geometrically: two lines have this angle of $\theta=\ldots$ between them is the same as saying that the lines are $\ldots$ relative to each other.

Your know the slope of the given line, viz. $m_1=-\frac{1}{\sqrt{3}}$. The slope of the tangent line is $m_2=\frac{dy}{dt}=\cot\left(\frac{1}{2}t+\frac{\pi}{4}\right)$. According to the conclusion in the last paragraph, these slopes must be $\ldots$ — that's how you can set up an equation to solve for $t$.

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