A) 4cos^2 x = 3-4cosx b)2sin^2 x + 3cosx -3 = 0

Question

A) 4cos^2 x = 3-4cosx
b)2sin^2 x + 3cosx -3 = 0

Answer 1

А
4сos²x+4cosx-3=0
cosx=a
4a²+4a-3=0
D=16+48=64
a1=(-4-8)/8=-1,5⇒cosx=-1,5<-1 нет решения<br>a2=(-4+8)/8=1/2⇒cosx=1/2⇒x=+-π/3+2πn,n∈z
b
2-2cos²x+3cosx-3=0
cosx=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πn,n∈z
a2=(3+1)/4=1⇒cosx=1⇒x=2πk,k∈z