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

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=2пn   или      x=+-arccos1/7+2пm

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