 # Number of solution in [ 0 , 2 &#x03C0;<!-- π --> ] satisfying the equation 8 sin Karina Trujillo 2022-06-26 Answered
Number of solution in $\left[0,2\pi \right]$ satisfying the equation
$8\mathrm{sin}\left(x\right)=\frac{\sqrt{3}}{\mathrm{cos}\left(x\right)}+\frac{1}{\mathrm{sin}\left(x\right)}$
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Multiply through by $\mathrm{sin}x\mathrm{cos}x$. We get
$8\mathrm{cos}x\left(1-{\mathrm{cos}}^{2}x\right)=\sqrt{3}\mathrm{sin}x+\mathrm{cos}x.$
Recall that $\mathrm{cos}3x=4{\mathrm{cos}}^{3}x-3\mathrm{cos}x$. So the left-hand side becomes $8\mathrm{cos}x-2\left(\mathrm{cos}3x+3\mathrm{cos}x\right)$, that is, $2\mathrm{cos}x-2\mathrm{cos}3x$. Thus our equation can be rewritten as
$\mathrm{cos}3x=\frac{1}{2}\mathrm{cos}x-\frac{\sqrt{3}}{2}\mathrm{sin}x.$
The right-hand side is equal to $\mathrm{cos}\left(x+\pi /3\right)$, so our equation simplifies to
$\mathrm{cos}3x=\mathrm{cos}\left(x+\pi /3\right).$
To finish, recall that $\mathrm{cos}s=\mathrm{cos}t$ if and only if $s=±t+2k\pi$ for some integer k.

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