;): Jeju skąd te zadanie jest?
cos3x = 4cos
3x − 3cosx
4cosx(cos
2 − 1) = 0
cosx(cosx − 1)(cosx + 1) = 0 x∊R
cosx = 0 ⋁ cosx = 1 ⋁ cosx = −1
| | π | |
x = |
| + kπ k∊C ⋁ x = 2kπ k∊C ⋁ x = π + 2kπ |
| | 2 | |
sin4x = sin(2x + 2x) =
sin2xcos2x + cosx2xsin2x =
2sin2xcos2x
2sin2xcos2x = 0 x∊<0,2π>
sin2x = 0
| | π | | 3 | |
x = 0 ⋁ x = |
| ⋁ x = π ⋁ x = |
| π ⋁ x = 2π |
| | 2 | | 2 | |
cos2x = 0
| | π | | 3 | | 5 | | 7 | |
x = |
| ⋁ x = |
| π ⋁ x = |
| π ⋁ x = |
| π |
| | 4 | | 4 | | 4 | | 4 | |
| | k | |
x∊{ |
| π ⋀ k∊{0,1,2,3,4,5,6,7,8}} |
| | 4 | |
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