Sarah-Louise Prince

2022-02-27

Solve equation
${\left(8{\mathrm{cos}}^{3}x+1\right)}^{3}=162\mathrm{cos}x-27$

Kathryn Duggan

This approach is based off knowing that the answers are . Another equation whose solutions are exactly those values, is $2\mathrm{cos}3x+1=0$, so we'd look to factorize that out.
We know that $2\mathrm{cos}3x+1=8{\mathrm{cos}}^{3}x-6\mathrm{cos}x+1$, so
$\left(8{\mathrm{cos}}^{3}x+1\right)=\left(2\mathrm{cos}3x+1+6\mathrm{cos}x\right)$. Henceforth, let $w=2\mathrm{cos}3x+1$
We have ${\left(w+6\mathrm{cos}x\right)}^{3}=162\mathrm{cos}x-27$, or that ${w}^{3}+3{w}^{26}\mathrm{cos}x+3w{\left(6\mathrm{cos}x\right)}^{2}+216{\mathrm{cos}}^{3}x=162\mathrm{cos}x-27=-27w$
hence ${w}^{3}+18{w}^{2}\mathrm{cos}x+108w{\mathrm{cos}}^{2}x+27w=0$
This factorizes into $0=w\left({w}^{2}+18w\mathrm{cos}x+108{\mathrm{cos}}^{2}x+27\right)$
$=w\left[{\left(w+9\mathrm{cos}x\right)}^{2}+27{\mathrm{cos}}^{2}x+27\right]$
It is clear that the other term is strictly positive, so has no solution for x (regardless of what w is), so we must have $w=0$. Hence, the solutions are

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