sprawdzenie
Monika: | | π | |
sinx(tgx+sinx)−cosx(tgx+sinx)=0 df=R\{ |
| +kπ} |
| | 2 | |
(tgx+sinx)(sinx−cosx)=0
tgx+sinx=0 sinx=sin(0,5π−x)
| sinxcosx+sinx | |
| =0 v x=0,5π−x+2kπ v x=π−0,5π+x+2kπ |
| cosx | |
tgx*(cosx+1)=0 v
x=0,25π+2kπ v (0=0,5π+2kπ)= sprzeczne
tgx=0 v cosx=−1
x=kπ ( x=−0,5π+2kπ v x=0,5π+2kπ)−sprzeczne
