Math Help - solve for x

1. solve for x

if -180<=x<=180 solve
6-10cosx=3sin^2x

2. Originally Posted by rionariona
if -180<=x<=180 solve
6-10cosx=3sin^2x
$\sin^2(x) = 1 - \cos^2(x)$

$6-10 \cos(x) = 3 - 3\cos^2(x)$ which is a quadratic in $cos(x)$

Put into the form $ax^2+bx+c=0$ and solve using the quadratic formula