Blericker74

2022-07-08

Solve
$6{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=5$
$0\le x\le {360}^{\circ }$

Jamarcus Shields

$6{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=5$
$6{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=5\cdot 1$
$6{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=5\cdot \left({\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)\right)$
$6{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=5{\mathrm{sin}}^{2}\left(x\right)+5{\mathrm{cos}}^{2}\left(x\right)$
${\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-6{\mathrm{cos}}^{2}\left(x\right)=0$
${\mathrm{tan}}^{2}\left(x\right)+\mathrm{tan}\left(x\right)-6=0$
$\mathrm{tan}\left(x\right)=2$
$x=\mathrm{arctan}2+\pi n,n\in \mathbb{Z}$
$x=-\mathrm{arctan}3+\pi k,k\in \mathbb{Z}$

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