Sin^2x+2cos^2x-5cosx-7=0

$sin ^{2} x+2cos ^{2} x-5cosx-7=0$
$1-cos ^{2} x+2cos ^{2} x-5cosx-7=0$
$cos ^{2} x-5cosx-6=0$
$cosx=t$
$t^{2} -5t-6=0$
D = (-5)² - 4·1·(-6) = 25 + 24 = 49
$sqrt{D} =7$

$t_{1} = frac{5+7}{2} = frac{12}{2} =6$

$t_{2} = frac{5-7}{2} = frac{-2}{2}=-1$

$left { {{cosx=6} atop {cosx=-1}} right.$

Ø
$x= pi +2 pi k,$ k∈Z