[(sinx+sin(3x)]+sin(2x)= [cosx+cos(3x)]+cos(2x)
2sin(2x)*cosx+sin(2x)= 2cos(2x)*cosx+cos(2x)
2sin(2x)*cosx+sin(2x)−2cos(2x)*cosx−cos(2x)=0
2sin(2x)*cosx−2cos(2x)*cosx + sin(2x)−cos(2x)=0
2sin(2x)*cosx−2cos(2x)*cosx + 2sinx*cosx −cos2+sin2x=0
2sin(2x)*cosx−2cos(2x)*cosx+2sinxcosx−2cos2x+1 =0
Z tego wyciagne cosx przed nawias
cosx[2sin(2x)−2cos(2x)+2sinx−2cosx]+1=0
Ale to mi wyszedl kosmos
| 1 | ||
cosx=− | sin2x=cos2x | |
| 2 |
dziekuje CI
Dokoncze
cosx= −0,5
| −π | ||
cosx= cos | ||
| 3 |
| π | ||
x=±(− | )+2kπ | |
| 3 |
| π | ||
sin(2x)= sin( | −2x) | |
| 2 |
| π | ||
2x= | −2x+2kπ lub 2x= (π−(π}{2}−2x))+2kπ | |
| 2 |
| π | ||
4x= | +2kπ | |
| 2 |
| π | 1 | π | π | |||||
x= | + | kπ lub 2x= | +2x+2kπ to dostane 0= | +2kπ (jak mam to | ||||
| 8 | 2 | 2 | 2 |
| 1 | ||
cosx=− | ||
| 2 |
| π | π | |||
x= | +π+2kπ lub x=− | +π+2kπ | ||
| 3 | 3 |
| 4π | 2π | |||
x= | +2kπ lub x= | +2kπ | ||
| 3 | 3 |
| 2π | π | |||
x= | (3n±1) x= | (4n+1) | ||
| 3 | 8 |
| √2 | ||
sin(2x)−cos(2x)=0 /* | ||
| 2 |
| √2 | √2 | ||
sin(2x)− | cos(2x)=0 | ||
| 2 | 2 |
| π | π | |||
cos | cos(2x)−sin | sin(2x)=0 | ||
| 4 | 4 |
| π | ||
cos( | +2x)=0 | |
| 4 |
| π | π | ||
+2x= | +kπ | ||
| 4 | 2 |
| 3π | ||
2x= | +kπ | |
| 4 |
| π | 1 | π | ||||
x= | + | kπ= | (4k+1) | |||
| 8 | 2 | 8 |
| π | ||
sin(2x) = sin( | − 2x) | |
| 2 |
| π | ||
2x = | − 2x + 2kπ | |
| 2 |
| π | ||
4x = | + 2kπ | |
| 2 |
| π | 1 | |||
x = | + | kπ | ||
| 8 | 2 |