Equation $(8\cos^3x+1)^3=162\cos x-27$

Question

Solve equation $$(8\cos^3x+1)^3=162\cos x-27$$

I saw this equation before 5 month, and I couldn't solve it. This isn't homework, etc. (I don't do stuff like this anymore). I am just curious.

Answer 1

We use herein the power-reduction formula $$\cos^3\theta=\frac{3\cos\theta+\cos 3\theta}4\tag{$\clubsuit$}$$ and the identity $$a^3-b^3=(a-b)\left((a+b)^2-ab\right).\tag{$\heartsuit$}$$

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Observe that by $(\clubsuit),$ we have $$8\cos^3x=2(3\cos x+\cos3x)=6\cos x+2\cos3x,$$ so $$\begin{align}162\cos x-27 &= 27(6\cos x-1)\\ &= 27(8\cos^3x-2\cos3x-1)\\ &= 27(8\cos^3x)-27(2\cos3x+1)\\ &= (6\cos x)^3-27(2\cos3x+1)\\ &= (8\cos^3x-2\cos3x)^3-27(2\cos3x+1).\end{align}$$ Thus, by $(\heartsuit),$ we find $$\begin{align}(8\cos^3x+1)^3-(162\cos x-27) &= (8\cos^3x+1)^3-(8\cos^3x-2\cos3x)^3+27(2\cos3x+1)\\ &= (2\cos3x+1)f(x),\end{align}$$ where $$f(x)=\left((8\cos^3x+1)+(8\cos^3x-2\cos3x)\right)^2-(8\cos^3x+1)(8\cos^3x-2\cos3x)+27.$$ Thus, we must find all $x$ such that $2\cos3x+1=0$ or $f(x)=0$. I leave the first task to you, and will show in the next section that $f(x)>0$ for all real $x$, so that our solutions are precisely those $x$ such that $2\cos3x+1=0$.

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Now, observe that $$\begin{align}f(x) &= \left(16\cos^3x+1-2\cos3x\right)^2-(8\cos^3x+1)(8\cos^3x-2\cos3x)+27\\ &= \bigl(4(3\cos x+\cos3x)+1-2\cos3x\bigr)^2-\bigl(2(3\cos x+\cos3x)+1\bigr)(6\cos x)+27\\ &= \bigl(12\cos x+2\cos3x+1\bigr)^2-(6\cos x+2\cos3x+1)(6\cos x)+27\\ &= \bigl(12\cos x+2\cos3x+1\bigr)^2-36\cos^2x-12\cos x\cos3x-6\cos x+27,\end{align}$$ so since $$\begin{align}\bigl(12\cos x+2\cos3x+1\bigr)^2 &= \bigl(12\cos x+2\cos3x+1\bigr)\bigl(12\cos x+2\cos3x+1\bigr)\\ &= 144\cos^2x+48\cos x\cos3x+24\cos x+4\cos^23x+4\cos3x+1,\end{align}$$ then $$f(x)=108\cos^2x+36\cos x\cos3x+18\cos x+4\cos^23x+4\cos3x+28.$$ By $(\clubsuit),$ we find that $$\cos3x=4\cos^3x-3\cos x,$$ so $$\begin{align}4\cos^23x+4\cos3x &= 4(4\cos^3x-3\cos x)^2+4(4\cos^3x-3\cos x)\\ &= 64\cos^6x-96\cos^4x+36\cos^2x+16\cos^3x-12\cos x\end{align}$$ and $$36\cos x\cos3x=144\cos^4x-108\cos^2x,$$ whence $$f(x)=64\cos^6x+48\cos^4x+16\cos^3x+36\cos^2x+6\cos x+28.$$ Consider the polynomial $$p(t)=64t^6+48t^4+16t^3+36t^2+6t+28.$$ Note that if $-1\le t,$ then $$\begin{align}p(t) &= 64|t|^6+48|t|^4+16|t|^2t+36|t|^2+6t+28\\ &\ge 64|t|^6+48|t|^4-16|t|^2+36|t|^2-6+28\\ &= 64|t|^6+48|t|^4+20|t|^2+22\\ &>0.\end{align}$$ Since $-1\le\cos x\le1$ for all real $x$, and since $f(x)=p(\cos x)$, it follows that $f(x)>0$ for all real $x$, as desired.

Answer 2

This approach is based off knowing that the answers are $ \frac {2 \pi}{9} , \frac {4 \pi }{9} , \frac { 8 \pi }{9} $. Another equation whose solutions are exactly those values, is $ 2 \cos 3 x + 1 = 0$, so we'd look to factorize that out.

We know that $ 2 \cos 3x + 1 = 8\cos^3 - 6 \cos x + 1$, so $ (8 \cos^3 x + 1) = (2 \cos 3x + 1 + 6 \cos x)$. Henceforth, let $w = 2 \cos 3x + 1$.

We have $ ( w + 6 \cos x) ^3 = 162 \cos x - 27$, or that $w^3 + 3 w^2 6\cos x + 3 w (6 \cos x)^2 + 216 \cos^3 x = 162 \cos x - 27$.

Continue to focus on factorizing out $w$, we get $w^3 + 18 w^2 \cos x + 108 w \cos^2x = - 216 \cos^3 x + 162 \cos x - 27 = - 27 w$,

hence $w^3 + 18w^2 \cos x + 108 w \cos^2x + 27 w = 0$.

This factorizes into $0 = w ( w^2 + 18 w \cos x + 108 \cos^2 x + 27) = w [ (w + 9 \cos x)^2 + 27 \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 $x = \frac {2\pi}{9}, \frac { 4\pi}{9}, \frac {8\pi}{9}$.