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Use an expression for $\frac{\mathrm{sin}\left(5\theta \right)}{\mathrm{sin}\left(\theta \right)}$ to find the roots of the equation ${x}^{4}-3{x}^{2}+1=0$ in trigonometric form
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rynosluv101swv2s
HINT:
Using Prosthaphaeresis Formula,
$\mathrm{sin}5x-\mathrm{sin}x=2\mathrm{sin}2x\mathrm{cos}3x=4\mathrm{sin}x\mathrm{cos}x\mathrm{cos}3x$
If $\mathrm{sin}x\ne 0,$,
$\frac{\mathrm{sin}5x}{\mathrm{sin}x}-1=4\mathrm{cos}x\mathrm{cos}3x=4\mathrm{cos}x\left(4{\mathrm{cos}}^{3}x-3\mathrm{cos}x\right)=\left(4{\mathrm{cos}}^{2}x{\right)}^{2}-3\left(4{\mathrm{cos}}^{2}x\right)$
OR replace ${\mathrm{sin}}^{2}x$ with $1-{\mathrm{cos}}^{2}x$ in your
$5{\mathrm{cos}}^{4}x-10{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x$
Now if $\mathrm{sin}5x=0,5x=n\pi$ where n is any integer
$x=\frac{n\pi }{5}$ where $n\equiv 0,±1,±2\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}5\right)$
So, the roots of $\frac{\mathrm{sin}5x}{\mathrm{sin}x}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=\frac{n\pi }{5}$ where $n\equiv ±1,±2\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}5\right)$
But
$\frac{\mathrm{sin}5x}{\mathrm{sin}x}=\left(4{\mathrm{cos}}^{2}x{\right)}^{2}-3\left(4{\mathrm{cos}}^{2}x\right)+1$