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

2

Ответы и объяснения

2014-02-28T08:17:33+00:00
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

2014-02-28T08:38:59+00:00
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