6(cosx)^2 + cos3x = cos x <--- cos3x = 4(cosx)^3 -3cosx</p>
6(cosx)^2 + 4(cosx)^3 -3cosx - cos x =0
4(cosx)^3 +6(cosx)^2 -4cosx =0 <----делим на 2 <----- cosx выносим за скобку</p>
cosx (2(cosx)^2 +3(cosx) -2) =0
cosx = 0 ; cosx =cos(π/2)
x =πn - π/2 , n € Z
или
2(cosx)^2 +3(cosx) -2 =0 <------cosx=t</p>
2t^2 +3t - 2=0
D=9 +16 =25 ; √D =5
t = (-3 +/- 5) /4
t1 = -2 ; не подходит, т к cosx > 0
t2 = 1/2 ; cosx =1/2 = cos (π/3)
x =2πn - π/3 , n € Z
x =2πn + π/3 , n € Z
ОТВЕТ
x =πn - π/2 , n € Z
x =2πn - π/3 , n € Z
x =2πn + π/3 , n € Z