Question

3 cos2x - 14 cosx + 7 = 0 Решите уравнение

Answer 1

3(2cos²x - 1) - 14cosx + 7 = 0
6cos²x - 3 - 14cosx + 7 = 0
6cos²x - 14cosx + 4 = 0
3cos²x - 7cosx + 2 = 0
D = 49 - 24 = 25
cosx = (7+5)/2 = 2 >1 - не подходит
cosx = (7-5)/6 = 1/3
x = +-arccos(1/3) + 2πk, k∈Z

Answer 2

$3 cos2x - 14 cosx + 7 = 0$
$3(2cos^2x-1)-14cosx+7=0$
$6cos^2x-3-14cosx+7=0$
$6cos^2x-14cosx+4=0$
$3cos^2x-7cosx+2=0$
Замена: $cosx=a,$   $|a| leq 1$
$3a^2-7a+2=0$
$D=(-7)^2-4*3*2=25$
$a_1= frac{7+5}{6} =2$  - не удовлетв.
$a_2= frac{7-5}{6}= frac{1}{3}$
$cosx =frac{1}{3}$
$x=бarccos frac{1}{3} +2 pi n,$ $n$ ∈ $Z$