Aliyah Mendez

2022-04-07

find the x: $\left(\mathrm{cos}3x+\mathrm{cos}4x\right)\left(\mathrm{cos}3x+\mathrm{cos}x\right)=\frac{1}{4}$

gil001q4wq

Let $\mathrm{cos}x=t$.
Thus, we get
$\left(4{t}^{3}-3t+8{t}^{4}-8{t}^{2}+1\right)\left(4{t}^{3}-3t+t\right)=\frac{1}{4}$
or
$\left(2t+1\right)\left(4{t}^{2}+2t-1\right)\left(16{t}^{4}-8{t}^{3}-16{t}^{2}+8t+1\right)=0$
$16{t}^{4}-8{t}^{3}-16{t}^{2}+8t+1=0$
we can solve by the following way
${\left(4{t}^{2}-t-\frac{3}{2}\right)}^{2}-5{\left(t-\frac{1}{2}\right)}^{2}=0.$
I think the rest is smooth.
I think the following way a bit of better.
After using of your work we need to solve
$16\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}\mathrm{cos}\left\{x\right\}\mathrm{cos}2x\mathrm{cos}\frac{7x}{2}=\mathrm{sin}\frac{x}{2}$
or
$2\mathrm{sin}4x\mathrm{cos}\frac{7x}{2}=\mathrm{sin}\frac{x}{2}$
or
$\mathrm{sin}\frac{15x}{2}+\mathrm{sin}\frac{x}{2}=\mathrm{sin}\frac{x}{2}$
or
$\mathrm{sin}\frac{15x}{2}=0$
Now, we need to delete all roots of $\mathrm{sin}\frac{x}{2}=0$ and to write the answer.

Substituting $\mathrm{cos}\left(t\right)=\frac{{e}^{it}+{e}^{-it}}{2}$ and letting $y={e}^{ix}$ , you get:
$\left(\mathrm{cos}\left(4x\right)+\mathrm{cos}\left(3x\right)\right)\left(\mathrm{cos}\left(3x\right)+\mathrm{cos}\left(x\right)\right)=\frac{1}{4}$
$\left({y}^{4}+{y}^{-4}+{y}^{3}+{y}^{-3}\right)\left({y}^{3}+{y}^{-3}+{y}^{1}+{y}^{-1}\right)=1$
$\left(\sum _{k=-7}^{7}{y}^{k}\right)+1=1$
$\sum _{k=0}^{14}{y}^{k}=0$
$\frac{{y}^{15}-1}{y-1}=0$
So ${e}^{15ix}=1$ and ${e}^{ix}\ne 1$, leaving the solutions being $x=\frac{2\pi n}{15}$ for $n\in \mathbb{Z}$ and 15∤n. I suppose it is just the coincidence of contrived problems that it worked out to be a geometric series.

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