# Solve 6 sin 2 </msup> &#x2061;<!-- ⁡ --> ( x ) + sin &#x2061;<!-- ⁡ -

Solve
$6{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)=5$
$0\le x\le {360}^{\circ }$
You can still ask an expert for help

## Want to know more about Trigonometry?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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