Решение задачи #59143
Решить уравнение cosx+2cos2x=1
cosx + 2cos2x = 1
cos2x = cos2x - sin2x
sin2x = 1 - cos2x
=>
cos2x = cos2x - 1 + cos2x = 2cos2x - 1
cosx + 2(2cos2x - 1) = 1
4cos2x + cosx - 3 = 0
t = cosx
4t2 + t - 3 = 0
D = 1 + 48 = 49
t1 = (-1 + 7):8 = 0,75
t2 = (-1 - 7):8 = -1
| [ | cosx = 0,75 |
| cosx = -1 |
| [ | x = ±arccos(0,75)+2πk |
| x = π + 2πk, |
где k ∈ Z.
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