Isolating the secant (with a coefficient of one) on one side of the equation gives,

\[\sec \left( {2t} \right) = \frac{1}{9}\]

At this point we can stop. We know that

\[\sec \theta \le - 1\hspace{0.25in}\,\,\,\,\,{\rm{or}}\hspace{0.25in}\,\,\,\,\sec \theta \ge 1\]

This means that it is impossible for secant to ever be \(\frac{1}{9}\) and so there will be no solution to this equation.

Note that if you didn’t recall the restrictions on secant the next step would have been to convert this to cosine so let’s do that.

\[\sec \left( {2t} \right) = \frac{1}{{\cos \left( {2t} \right)}} = \frac{1}{9}\hspace{0.50in} \Rightarrow \hspace{0.50in}\cos \left( {2t} \right) = 9\]

At this point we can note that \( - 1 \le \cos \theta \le 1\) and so again there is no way for cosine to be 9 and again we get that there will be no solution to this equation.