David Young

2021-12-31

What is general solution of $\frac{\mathrm{cos}5x\mathrm{cos}3x-\mathrm{sin}3x\mathrm{sin}x}{\mathrm{cos}2x}=1$
1) $\frac{k\pi }{3}$
2) $\frac{k\pi }{2}$
3) $\frac{2k\pi }{5}$
4) $\frac{2k\pi }{3}$
The numerator of the fraction is $\mathrm{cos}\left(5x+3x\right)$. so I should find general solution of $\mathrm{cos}8x=\mathrm{cos}2x$. I'm not sure how to do it, I can write $\mathrm{cos}8x$ in term of $\mathrm{cos}2x$:
$\mathrm{cos}8x=2{\mathrm{cos}}^{2}4x-1=2{\left(2{\mathrm{cos}}^{2}2x-1\right)}^{2}-1$
After substituting it in the equation and using $\mathrm{cos}2x=t$ we have degree four equation

jean2098

$0=\mathrm{cos}5x\mathrm{cos}3x-\mathrm{sin}3x\mathrm{sin}x-\mathrm{cos}2x$
$=\mathrm{cos}5x\mathrm{cos}3x-\mathrm{sin}3x\mathrm{sin}x-\mathrm{cos}\left(5x-3x\right)$
$=\mathrm{sin}3x\mathrm{sin}5x-\mathrm{sin}3x\mathrm{sin}x$
$\to -2{\mathrm{sin}}^{2}3x\mathrm{cos}2x=0$
Note that $\mathrm{cos}2x\ne 0$ has to be included
$\to \mathrm{sin}3x=0$

William Appel

$\mathrm{cos}\left(5x\right)\mathrm{cos}\left(3x\right)-\mathrm{sin}\left(3x\right)\mathrm{sin}\left(x\right)=\mathrm{cos}\left(2x\right)$
$\frac{12}{\mathrm{cos}\left(2x\right)+\mathrm{cos}\left(8x\right)}-\frac{12}{\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(4x\right)}=\mathrm{cos}\left(2x\right)$
$\frac{12}{\mathrm{cos}\left(4x\right)+\mathrm{cos}\left(8x\right)}-\mathrm{cos}\left(2x\right)=0$
$\mathrm{cos}\left(8x\right)+\mathrm{cos}\left(4x\right)-2\mathrm{cos}\left(2x\right)=0$
$2\mathrm{cos}\left(2x\right)\mathrm{cos}\left(6x\right)-2\mathrm{cos}\left(2x\right)=0$
$\mathrm{cos}\left(2x\right)\left[\mathrm{cos}\left(6x\right)-1\right]=0$
$\mathrm{cos}\left(2x\right)=0\to x=±\frac{\pi }{4}+k\pi$
discarded because this solution will make zero the denominator of the original equation
$\mathrm{cos}\left(6x\right)=1\to x=\frac{k\pi }{3}$

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