Question

решить уравнение. 7sin^2x + 8cosx -8=0 отрезок [-п/2 ; п/2]

Answer 1

7-7cos²x+8cosx-8=0
cosx=a
7a²-8a+1=0
D=64-28=36
a1=(8-6)14=1/7⇒cosx=1/7
x=-arccos1/7+2πn U x=arccos1/7+2πn,n∈z
x=+-arccos1/7∈[-π/2;π/2]
a2=(8+6)/14=1⇒cosx=1⇒x=2πn,n∈z
x=0∈[-π/2;π/2]