[tex]9\cos(2t)=6\implies\cos(2t)=\dfrac23[/tex]
Using the fact that cos is 2π-periodic, we have
[tex]\cos(2t)=\dfrac23\implies2t=\cos^{-1}\left(\dfrac23\right)+2n\pi[/tex]
That is, [tex]\cos(\theta+2n\pi)=\cos\theta[/tex] for any [tex]\theta[/tex] and integer [tex]n[/tex].
[tex]\implies t=\dfrac12\cos^{-1}\left(\dfrac23\right)+n\pi[/tex]
We get 2 solutions in the interval [0, 2π] for [tex]n=0[/tex] and [tex]n=1[/tex],
[tex]t=\dfrac12\cos^{-1}\left(\dfrac23\right)\text{ and }t=\dfrac12\cos^{-1}\left(\dfrac23\right)+\pi[/tex]
