SOLUTION
Write out the equation given
[tex]\cos ^2x+2\cos x-3=0[/tex]
Let
[tex]\text{Cosx}=P[/tex]
Then by substitution, we obtain the equation
[tex]p^2+2p-3=0[/tex]
Solve the quadractic equation using factor method
[tex]\begin{gathered} p^2+3p-p-3=0 \\ p(p+3)-1(p+3)=0 \\ (p-1)(p+3)=0 \end{gathered}[/tex]
Then we have
[tex]\begin{gathered} p-1=0,p+3=0 \\ \text{Then} \\ p=1,p=-3 \end{gathered}[/tex]
Recall that
[tex]\cos x=p[/tex]
Hence
[tex]\begin{gathered} \text{when p=1} \\ \cos x=1 \\ \text{Then } \\ x=\cos ^{-1}(1)=0 \\ \text{hence } \\ x=0 \end{gathered}[/tex]
Similarly,
[tex]\begin{gathered} \text{When p=-3} \\ \cos x=-3 \\ x=\cos ^{-1}(-3) \\ x=no\text{ solution} \end{gathered}[/tex]
Therefore x=0 is the only valid solution on the given interval [0,2π).
Answer; x=0
