delirija7z

2022-07-10

Show analytically that 0 is the only zero of $\mathrm{sin}\left(2x\right)+2x$
I tried:
$0=\mathrm{sin}\left(2x\right)+2x⇔\phantom{\rule{0ex}{0ex}}0=2\mathrm{sin}x\mathrm{cos}x+2x⇔\phantom{\rule{0ex}{0ex}}0=2\left(\mathrm{sin}x\mathrm{cos}x+x\right)⇔\phantom{\rule{0ex}{0ex}}0=\mathrm{sin}x\sqrt{1-{\mathrm{sin}}^{2}x}+x⇔\phantom{\rule{0ex}{0ex}}0=\sqrt{{\mathrm{sin}}^{2}x\left(1-{\mathrm{sin}}^{2}x\right)}+x⇔\phantom{\rule{0ex}{0ex}}0=\sqrt{{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{4}x}+x⇔\phantom{\rule{0ex}{0ex}}???$

Expert

Hint: Consider the increasing/decreasing behavior of the function $f\left(x\right)=\mathrm{sin}\left(2x\right)+2x$
${f}^{\prime }\left(x\right)=2\mathrm{cos}\left(2x\right)+2$. Since the range of $\mathrm{cos}\theta$ is $\left[-1,1\right]$, we know that $2\mathrm{cos}\left(2x\right)+2\ge 0$, so f is (non-strictly) increasing for all real numbers.

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