Question

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

Answer 1

$sin ^{2} x+2cos ^{2} x-5cosx-7=0$
$1-cos ^{2} x+2cos ^{2} x-5cosx-7=0$
$cos ^{2} x-5cosx-6=0$
$cosx=t$
$t^{2} -5t-6=0$
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