Решите уравнение 7 sin² x + 8 cos x - 8 = 0 и найдите корни, принадлежащие отрезку - π на два; π на два

А) 7 (1-cos²x) + 8cosx-8=0

7-7cos²x+8cosx-8=0

7cos²x-8cosx+1=0

D=64-28=36

cosx = (8+6) / 14=1; cosx = (8-6) / 14=1/7

x1=2πk, k∈Z

x2=arccos1/7+2πk, k∈Z;

x3=-arccos1/7+2πk, k∈Z;

б) - π/2≤2πk≤π/2

-1/4≤k≤1/4

k=0

x1=2π0=0

-π/2≤arccos1/7+2πk≤π/2;

-π/2-arccos1/7≤2πk≤π/2-arccos1/7

-1/4 - (arccos1/7) / 2π≤k≤1/4 - (arccos1/7) / 2π

k=0

x2=arccos1/7+2π0=arccos1/7;

-π/2≤-arccos1/7+2πk≤π/2;

-π/2+arccos1/7≤2πk≤π/2+arccos1/7

-1/4 + (arccos1/7) / 2π≤k≤1/4 + (arccos1/7) / 2π

k=0

x3=-arccos1/7+2π0=-arccos1/7

Ответ: а) x1=2πk, k∈Z; x2=arccos1/7+2πk, k∈Z; x3=-arccos1/7+2πk, k∈Z;

б) x1=0; x2=arccos1/7; x2=-arccos1/7

sin^2x=1-cos^2x

7 (1-cos^2x) + 8cos^2x-8=0

7-7cos^2x+8cosx-8=0

7cos^2x-8cosx+1=0

cosx=t

7t^2-8t^2+1=0

D=64-28=36

t1=1

t2=1/7

cosx=1

cosx=1/7

x=2 пn или x=+-arccos1/7+2 пm