równanie, sinus, cosinus 3x

[sea_of_tears]

Rozwiąż :

\(\displaystyle{ sin(x)-cos(3x)=0}\)

[milaR]

można skorzystać z tego ,że:
\(\displaystyle{ cos( \frac{\Pi}{2}+x)=-sinx}\)
czyli
\(\displaystyle{ sinx=-cos( \frac{\Pi}{2}+x)}\)
\(\displaystyle{ -cos( \frac{\Pi}{2}+x)-cos(3x)=0}\)
\(\displaystyle{ cos( \frac{\Pi}{2}+x)+cos(3x)=0}\)
\(\displaystyle{ 2cos \frac{x+ \frac{\Pi}{2}+3x }{2} cos \frac{x+ \frac{\Pi}{2}-3x }{2} =0}\)
\(\displaystyle{ cos(2x+ \frac{\Pi}{4})=0 cos( \frac{\Pi}{4}-x)=0}\)
\(\displaystyle{ 2x+ \frac{\Pi}{4}= \frac{\Pi}{2}+k\Pi \frac{\Pi}{4}-x= \frac{\Pi}{2}+k\Pi}\)
\(\displaystyle{ 2x= \frac{\Pi}{4} +k\Pi -x=\frac{\Pi}{4} +k\Pi}\)
\(\displaystyle{ x= \frac{\Pi}{8} + \frac{k\Pi}{2} x=- \frac{\Pi}{4}-k\Pi}\)

[ Dodano: 16 Lutego 2008, 18:28 ]
gdzie \(\displaystyle{ k C}\)