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Section 1.6 : Solving Trig Equations with Calculators, Part II

10. Find all the solutions to \(\displaystyle \left( {2t - 3} \right)\tan \left( {\frac{{6t}}{{11}}} \right) = 15 - 10t\). 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.
Start Solution

This one may be a little trickier to factor than the others in this section, but it can be factored. First get everything on one side of the equation and then notice that we can factor out a \(2t - 3\) from the equation as follows,

\[\begin{align*}\left( {2t - 3} \right)\tan \left( {\frac{{6t}}{{11}}} \right) + 10t - 15 & = 0\\ \left( {2t - 3} \right)\tan \left( {\frac{{6t}}{{11}}} \right) + 5\left( {2t - 3} \right) & = 0\\ \left( {2t - 3} \right)\left( {\tan \left( {\frac{{6t}}{{11}}} \right) + 5} \right) & = 0\end{align*}\]

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

\[2t - 3 = 0\hspace{0.25in}{\rm{OR}}\hspace{0.25in}\tan \left( {\frac{{6t}}{{11}}} \right) + 5 = 0\]

Be careful with this type of equation to not make the mistake of just canceling the \(2t - 3\) from both sides. Had you done that you would have missed the solution from the first equation.

When solving equations it is important to remember that you can’t cancel anything from both sides unless you know for a fact that what you are canceling will never be zero.

Hint : Solve each of these two equations to attain all the solutions to the original equation.
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Solving the first equation gives,

\[t = \frac{3}{2}\]

Now we can move onto the second equation and note that this equation is similar to equations solved in the previous 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.

First, isolating the tangent gives,

\[\tan \left( {\frac{{6t}}{{11}}} \right) = - 5\]

Using our calculator we get,

\[\frac{{6t}}{{11}} = {\tan ^{ - 1}}\left( { - 5} \right) = - 1.3734\]

From our knowledge of the unit circle we can see that a positive angle that corresponds to this angle is \(2\pi - 1.3734 = 4.9098\). Either these angles can be used here but we’ll use the positive angle to avoid the possibility of losing the minus sign. Also, the second angle in the range \(\left[ {0,2\pi } \right]\) is \(\pi + \left( { - 1.3734} \right) = 1.7682\).

Finally, the solutions to this equation are,

\[\begin{align*}\frac{{6t}}{{11}} & = 1.7682 + 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in}\frac{{6t}}{{11}} & = 4.9098 + 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \\t & = 3.2417 + \frac{{11\pi n}}{3} & \hspace{0.25in}{\rm{OR}}\hspace{0.45in}t & = 9.0013 + \frac{{11\pi n}}{3} & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \end{align*}\]

Putting all of this together gives the following set of solutions.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{3}{2}, \hspace{0.25in} t = 3.2417 + \frac{{11\pi n}}{3}, \hspace{0.25in} {\rm{OR}} \hspace{0.25in} t = 9.0013 + \frac{{11\pi n}}{3}\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }}\]

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.