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Home / Calculus I / Review / Trig Equations with Calculators, Part I
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Section 1.5 : Solving Trig Equations with Calculators, Part I

9. Find the solution(s) to \(5 - 14\tan \left( {8x} \right) = 30\) that are in \(\left[ { - 1,1} \right]\). Use at least 4 decimal places in your work.

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Hint : Find all the solutions to the equation without regard to the given interval. The first step in this process is to isolate the tangent (with a coefficient of one) on one side of the equation.
Start Solution

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

\[\tan \left( {8x} \right) = - \frac{{25}}{{14}}\]
Hint : Using a calculator and your knowledge of solving trig equations involving tangents to determine all the angles in the range \(\left[ {0,2\pi } \right]\) for which tangent will have this value.
Show Step 2

First, using our calculator we can see that,

\[8x = {\tan ^{ - 1}}\left( { - \frac{{25}}{{14}}} \right) = - 1.0603\]

As we discussed in Example 5 of this section the second angle for equations involving tangent will always be the \(\pi \) plus the first angle. Therefore, \(\pi + \left( { - 1.0603} \right) = 2.0813\) will be the second angle.

Hint : Using the two angles above write down all the angles for which tangent will have this value and use these to write down all the solutions to the equation.
Show Step 3

From the discussion in the notes for this section we know that once we have these two angles we can get all possible angles by simply adding “\( + \,2\pi n\) for \(n = 0, \pm 1, \pm 2, \ldots \)” onto each of these.

This then means that we must have,

\[8x = - 1.0603 + 2\pi n\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}8x = 2.0813 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \]

Finally, to get all the solutions to the equation all we need to do is divide both sides by 8 and we’ll convert everything to decimals to help with the final step.

\[\begin{align*}x & = - 0.1325 + \frac{{\pi n}}{4} & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} & x = 0.2602 + \frac{{\pi n}}{4}\hspace{0.25in} & n = 0, \pm 1, \pm 2, \ldots \\ & = - 0.1325 + 0.7854n & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} & \hspace{0.10in}= 0.2602 + 0.7854n\hspace{0.25in} & n = 0, \pm 1, \pm 2, \ldots \end{align*}\]
Hint : Now all we need to do is plug in values of \(n\) to determine which solutions will actually fall in the given interval.
Show Step 4

Now let’s find all the solutions.

\[\begin{array}{lclcl}{n = - 1:}&{x = - 0.9179}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{x = - 0.5252}\\{n = 0:}&{x = - 0.1325\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{x = 0.2602}\\{n = 1:\,}&{x = 0.6529\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{x = \require{cancel} \xcancel{{1.0456}} > 1}\end{array}\]

Notice that with each increase in \(n\) we were really just adding/subtracting another 0.7854 from the previous results. A quick inspection of the results above will quickly show us that we don’t need to go any farther and we won’t bother with any other values of \(n\). Also, as we’ve seen in this problem it is completely possible for only one of the solutions from a given interval to be in the given interval so don’t worry about that when it happens.

So, it looks like we have the five solutions to this equation in the given interval.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 0.9179,\,\, - 0.5252,\,\, - 0.1325,\,\,0.2602,\,\,0.6529}}\]

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.