Sarah-Louise Prince

2022-02-27

Solve equation

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

Kathryn Duggan

Beginner2022-02-28Added 7 answers

This approach is based off knowing that the answers are $\frac{2\pi}{9},\text{}\frac{4\pi}{9},\text{}\frac{8\pi}{9}$ . 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

$(8{\mathrm{cos}}^{3}x+1)=(2\mathrm{cos}3x+1+6\mathrm{cos}x)$ . Henceforth, let $w=2\mathrm{cos}3x+1$

We have${(w+6\mathrm{cos}x)}^{3}=162\mathrm{cos}x-27$ , 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$0=w({w}^{2}+18w\mathrm{cos}x+108{\mathrm{cos}}^{2}x+27)$

$=w[{(w+9\mathrm{cos}x)}^{2}+27{\mathrm{cos}}^{2}x+27]$

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 $x=\frac{2\pi}{9},\text{}\frac{4\pi}{9},\text{}\frac{8\pi}{9}$

We know that

We have

hence

This factorizes into

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

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