Решение задачи #59163
Решить уравнение сos4x+2cos²х=1
cos4x = (cos2x)2 - (sin2x)2
cos2x = cos2x - sin2x = 2cos2x − 1
sin2x = 2sinx*cosx
sin2x = 1 - cos2x
сos4x + 2cos2х = 1
(cos2x - sin2x)2 - (2sinx*cosx)2 + 2cos2х = 1
(2cos2x - 1)2 + (2cos2x - 1) - 4cos2х(1 - cos2x) = 0
4cos4x - 4cos2x + 1 + 2cos2x - 1 - 4cos2x + 4cos4x = 0
cos2x = t
4t2 - 4t + 1 + 2t - 1 - 4t + 4t2 = 0
8t2-6t = 0
| [ | t = 0 |
| 8t - 6 = 0 |
| [ | t = 0 |
| t = 0,75 |
| [ | cos2x = 0 |
| cos2x = 0,75 |
| [ | 0,5cos2x + 0,5 = 0 |
| 0,5cos2x + 0,5 = 0,75 |
| [ | cos2x = -1 |
| cos2x = 0,5 |
| 108px;">[ | 2x = -π + 2πk |
| 2x = π/3 + 2πk | |
| 2x = -2π/3 + 2πk |
| [ | x = -0,5π + πk |
| x = π/6 + πk | |
| x = -π/3 + πk |
где k - любое целое число.
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