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

Talamancoeb
2021-12-17
Answered

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

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nghodlokl

Answered 2021-12-18
Author has **33** answers

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,

${\mathrm{cos}}^{2}\left(x\right)-(1-{\mathrm{cos}}^{2}\left(x\right))-\mathrm{cos}\left(x\right)=0$

$2{\mathrm{cos}}^{2}\left(x\right)-\mathrm{cos}\left(x\right)-1=0$

$(\mathrm{cos}\left(x\right)-1)(2\mathrm{cos}\left(x\right)+1)=0$

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 $[0;2\pi )$ . 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.

And we have

Now, get everything in terms of cosine:

Therefore,

Now, solve

Implies that

Implies that

Ethan Sanders

Answered 2021-12-19
Author has **35** answers

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