Talamancoeb

2021-12-17

Find the solution of $\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(x\right)=0$ in the interval $\left[0;2\pi \right)$

nghodlokl

We remember that $\mathrm{cos}\left(2x\right)={\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)$
And we have
${\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)-\mathrm{cos}\left(x\right)=0$
Now, get everything in terms of cosine:
${\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=1$
${\mathrm{sin}}^{2}\left(x\right)=1-{\mathrm{cos}}^{2}\left(x\right)$
Therefore,
${\mathrm{cos}}^{2}\left(x\right)-\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)-\mathrm{cos}\left(x\right)=0$
$2{\mathrm{cos}}^{2}\left(x\right)-\mathrm{cos}\left(x\right)-1=0$
$\left(\mathrm{cos}\left(x\right)-1\right)\left(2\mathrm{cos}\left(x\right)+1\right)=0$
Now, solve
$\mathrm{cos}\left(x\right)-1=0$
$2\mathrm{cos}\left(x\right)+1=0$
$\mathrm{cos}\left(x\right)=1$
Implies that $x=0$, as we're working in $\left[0;2\pi \right)$. And it's the only value yielding one for cosine.
$2\mathrm{cos}\left(x\right)=-1$
$\mathrm{cos}\left(x\right)=-\frac{1}{2}$
Implies that $x=\frac{2\pi }{3},\frac{4\pi }{3}$ as these values return $-\frac{1}{2}$ for cosine in the given interval.

Ethan Sanders

$\mathrm{cos}x=\frac{1}{2}⇔x=\frac{2\pi }{3};x=\frac{4\pi }{3}$
$\mathrm{cos}x=1⇔x=0$
$2\pi$ is not included in given interval

Do you have a similar question?