2021-12-30

How do I solve the following:

alkaholikd9

Not much different form Toby Maks

Cleveland Walters

$a+b\ge 2\sqrt{ab}$
$9{\mathrm{tan}}^{2}x+1{\mathrm{cot}}^{2}x\ge 2\sqrt{9{\mathrm{tan}}^{2}x×1{\mathrm{cot}}^{2}x}=6$
so
$\mathrm{cos}\left(12x\right)-5\mathrm{sin}\left(3x\right)\ge 6$
Implicit
$\mathrm{cos}12x-5\mathrm{sin}3x\le max\left\{\mathrm{cos}12x\right\}+max\left\{-5\mathrm{sin}3x\right\}\le 1+5$
and only possibilities :
$\left\{\begin{array}{c}\mathrm{cos}12x=1\\ \mathrm{sin}3x=-1\end{array}$

nick1337

The minimum value of $9{\mathrm{tan}}^{2}x+{\mathrm{cot}}^{2}x$ is $2\sqrt{9{\mathrm{tan}}^{2}x{\mathrm{cot}}^{2}x}=6$ by AM-GM for all real x, as Thus $\mathrm{cos}\left(12x\right)-\left(5\mathrm{sin}\left(3x\right)+9\mathrm{tan}2x+\mathrm{cot}2x\right)$ is bounded above by $g\left(x\right)=\mathrm{cos}\left(12x\right)-\left(5\mathrm{sin}\left(3x\right)+6\right)$
Now g(x) is again bounded above by 1-5(-1)-6=0. Hence you just need to solve g(x)=0, but as for all $x\in \mathbb{R}$, this implies:

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