Question

cos^2(x)-8cos(x)+7=0

Answer 1

cos²x - 8cosx + 7 = 0
cosx = t
t² - 8t + 7 = 0
D = (-8)² - 4· 1· 7 = 64 - 28 = 36
$sqrt{D}=6$

$t _{1}= frac{8+6}{2} = frac{14}{2}=7$

$t _{2} = frac{8-6}{2} =frac{2}{2}=1$

$left { {{cosx=7} atop {cosx=1}} right.$

Ø
$x=2 pi k,$ k∈ Z