rozwiąż równanie (roz)

[kojot-wsp]

\(\displaystyle{ sin^2x-3cosx-3=0}\)

[math questions]

\(\displaystyle{ sin^2x-3cosx-3=0}\)

\(\displaystyle{ 1-cos ^{2}x-3cosx-3=0\\ -cos ^{2}x-3cosx-2=0\\ cosx=t,\ gdzie\ cosx \in <-1;1>\\ -t ^{2}-3t-2=0\\ \Delta=1 \Rightarrow \sqrt{\Delta} =1}\)

\(\displaystyle{ t _{1}=-2 \vee t _{2}=-1}\)

\(\displaystyle{ cosx=-1 \Rightarrow x= \pi +2k \pi ,\ gdzie\ k \in C}\)

[ppolciaa17]

\(\displaystyle{ sin^{2}x= 1-cos^{2}x}\)
\(\displaystyle{ 1-cos^{2}-3cosx-3=0}\)
\(\displaystyle{ cos^{2}x+3cosx+2=0}\)

\(\displaystyle{ t=cosx \
t \in<-1;1>}\)
\(\displaystyle{ t^{2}+2t+t+2=0}\)
\(\displaystyle{ t(t+2)+(t+2)=0}\)
\(\displaystyle{ t=-2 \vee t=-1}\)

\(\displaystyle{ cosx=-1}\)
\(\displaystyle{ x= \pi+2k \pi}\)
\(\displaystyle{ k \in C}\)