5cos²x-6cosx+1=0 (-3/2П; П)

Question

5cos²x-6cosx+1=0 (-3/2П; П)

Answer 1

cosx=t

5t^2-6t+1=0

(3+-sqrt(9-5))/5=(3+-2)/5

t1=1/5

t2=1

cosx=1

x=2Пk

cosx=1/5

x=+-arccos(1/5)+2Пk

 

x1=-arccos(1/5)

x2=arccos(1/5)

x3=0