3 cos2x - 14 cosx + 7 = 0 Решите уравнение

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

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