Natali Belska
Просветленный
(36219)
1 год назад
cos 2x = 2*cos^2 - 1
=>
2*(2*cos^2 - 1) + 4*cos x - 1 = 0
4cos^2 x + 4*cos x - 3 = 0 -----> cos x = t
4*t^2 + 4t - 3 = 0
t(1,2) = [-4 + - V(16 + 4*4*3)]/2*4 = (-4 + - 8)/8 = - 1/2 + - 1
cos x = t1 = - 1/2 - 1 = (-1-2)/2 = - 3/2 - быть не может
cos x = t1 = - 1/2 +1 = 1/2 ----> x = ...