I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 1.5 : Solving Trig Equations with Calculators, Part I
12. Find the solution(s) to \(\displaystyle \frac{4}{3} = 1 + 3\sec \left( {2t} \right)\) that are in \(\left[ { - 4,6} \right]\). Use at least 4 decimal places in your work.
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.