<s>[INST] Point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and the $x$-axis How can I algebraically (without looking at the graph) find the point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and $x$-axis, in the interval $[0, \pi]$?
 [/INST] Notice, $$f(x)=\sin 2x+\cos 2x$$ $$\implies f(x)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos 2x+\frac{1}{\sqrt{2}}\sin 2x\right)$$ $$=\sqrt{2}\left(\cos 2x\cos\frac{\pi}{4}+\sin 2x\sin\frac{\pi}{4}\right)$$ $$=\sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)$$ Now, for the intersection of $f(x)$ with the x-axis, we have $f(x)=0$ $$\implies \sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)=0 $$ $$\implies \cos \left(2x-\frac{\pi}{4}\right)=0 $$ Writing the general solution as follows $$\implies 2x-\frac{\pi}{4}=\frac{(2n+1)\pi}{2}$$ $$\implies x=\frac{(2n+1)\pi}{4}+\frac{\pi}{8}$$ $$\implies x=\frac{(4n+3)\pi}{8}$$ Where, $n$ is any integer 
For given interval $[0, \pi]$, put $n=0$ & $n=1$ in the general solution, we get  $$\color{blue}{x\in \left\{\frac{3\pi}{8}, \frac{7\pi}{8}\right\}}$$
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