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Section 1.6 : Solving Trig Equations with Calculators, Part II

7. Find all the solutions to \(4{\csc ^2}\left( {1 - t} \right) + 6 = 25\csc \left( {1 - t} \right)\). Use at least 4 decimal places in your work.

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Hint : Factor the equation and using basic algebraic properties get two equations that can be dealt with using known techniques. If you’re not sure how to factor this think about how you would factor \(4{x^2} - 25x + 6 = 0\).
Start Solution

This equation may look very different from anything that we’ve ever been asked to factor, however it is something that we can factor. First think about factoring the following,

\[4{x^2} + 6 = 25x\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,4{x^2} - 25x + 6 = \left( {4x - 1} \right)\left( {x - 6} \right) = 0\]

If we can factor this algebraic equation then we can factor the given equation in exactly the same manner.

\[\begin{align*}4{\csc ^2}\left( {1 - t} \right) + 6 & = 25\csc \left( {1 - t} \right)\\ 4{\csc ^2}\left( {1 - t} \right) - 25\csc \left( {1 - t} \right) + 6 & = 0\\ \left( {4\csc \left( {1 - t} \right) - 1} \right)\left( {\csc \left( {1 - t} \right) - 6} \right) & = 0\end{align*}\]

Now, we have a product of two factors that equals zero and so by basic algebraic properties we know that we must have,

\[4\csc \left( {1 - t} \right) - 1 = 0\hspace{0.25in}{\rm{OR}}\hspace{0.25in}\csc \left( {1 - t} \right) - 6 = 0\]
Hint : Solve each of these two equations to attain all the solutions to the original equation.
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Each of these equations are similar to equations solved in the previous section and earlier in this section. Therefore, we will be assuming that you can recall the solution process for each and we will not be putting in as many details. If you are unsure of the process you should go back to the previous section and work some of the problems there before proceeding with the solution to this problem.

We’ll start with the first equation, isolate the cosecant and convert to an equation in terms of sine for easier solving. Doing this gives,

\[\csc \left( {1 - t} \right) = \frac{1}{4}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {1 - t} \right) = 4 > 1\]

We now know that there are now solutions to the first equation because we know \( - 1 \le \sin \theta \le 1\).

Now, let’s solve the second equation.

\[\csc \left( {1 - t} \right) = 6\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \left( {1 - t} \right) = \frac{1}{6}\]

Using our calculator we get,

\[1 - t = {\sin ^{ - 1}}\left( {\frac{1}{6}} \right) = 0.1674\]

A quick glance at a unit circle shows us that the second angle in the range \(\left[ {0,2\pi } \right]\) is \(\pi - 0.1674 = 2.9742\).

All the solutions to the second equation are then,

\[\begin{align*}1 - t & = 0.1674 + 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in}1 - t & = 2.9742 + 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \\t & = 0.8326 - 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.55in}t & = - 1.9742 - 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

Because we had not solutions to the first equation all the solutions to the original equation are then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0.8326 - 2\pi n \hspace{0.25in} {\rm{OR}} \hspace{0.25in} t = - 1.9742 - 2\pi n \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots }}\]

Do get too excited about the fact that we only got solutions from one of the two equations we got after factoring. This will happen on occasion and so we need to be ready for these cases when they happen.

If an interval had been given we would next proceed with plugging in values of \(n\) to determine which solutions fall in that interval. Since we were not given an interval this is as far as we can go.

Note that depending upon the amount of decimals you used here your answers may vary slightly from these due to round off error. Any differences should be slight and only appear around the 4th decimal place or so however.