| x | ||
Rozwiąż równanie sin2x=cos2x+cos | w przedziale <−2π; 2π> | |
| 2 |
| x | ||
sin2x−cos2x=cos | ||
| 2 |
| x | ||
−(cos2x−sin2x)=cos | ||
| 2 |
| x | ||
−cos2x=cos | ||
| 2 |
| x | ||
0=cos2x+cos | ||
| 2 |
| 5x | 3x | |||
2cos | cos | =0 | ||
| 4 | 4 |
| 5x | 3x | |||
cos | =0 v cos | =0 | ||
| 2 | 4 |
| 5x | π | 3x | π | ||||
= | +kπ v | = | +kπ | ||||
| 4 | 2 | 4 | 2 |
| 2π | 4kπ | 2π | 4kπ | |||||
x= | + | v x= | + | k∊ℤ | ||||
| 5 | 5 | 3 | 3 |
| −10π | ||
x1=−2π x= | ∉ D | |
| 3 |
| −6π | ||
x2= | x=−2π (już było) | |
| 5 |
| −2π | −2π | |||
x3= | x4= | |||
| 5 | 3 |
| 2π | 2π | |||
x5= | x6= | |||
| 5 | 3 |
| 6π | ||
x7= | x8=2π | |
| 5 |
| −6π | −2π | −2π | 2π | 2π | 6π | |||||||
Odp. {−2π, | , | , | , | , | , | , 2π} | ||||||
| 5 | 5 | 3 | 5 | 3 | 5 |
wolfram też tak mówi
https://www.wolframalpha.com/input/?i=sin%5E2%28x%29%3Dcos%5E2%28x%29%2Bcos%28x%2F2%29+and++-2pi+≤+x+≤+2pi
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