Section 1.5 : Solving Trig Equations with Calculators, Part I
5. Find the solution(s) to \(\displaystyle 9\cos \left( {\frac{{4z}}{9}} \right) + 21\sin \left( {\frac{{4z}}{9}} \right) = 0\) that are in \(\left[ { - 10,10} \right]\). Use at least 4 decimal places in your work.
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Because we’ve got both a sine and a cosine here it makes some sense to reduce this down to tangent. So, reducing to a tangent (with a coefficient of one) on one side of the equation gives,
\[\tan \left( {\frac{{4z}}{9}} \right) = - \frac{3}{7}\]First, using our calculator we can see that,
\[\frac{{4z}}{9} = {\tan ^{ - 1}}\left( { - \frac{3}{7}} \right) = - 0.4049\]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, \(\pi + \left( { - 0.4049} \right) = 2.7367\) will be the second angle.
Also, because it is very easy to lose track of minus signs we’ll use the fact that we know that any angle plus \(2\pi \) will give another angle whose terminal line is identical to the original angle to eliminate the minus sign on the first angle. So, another angle that will work for the first angle is \(2\pi + \left( { - 0.4049} \right) = 5.8783\). Note that there is nothing wrong with using the negative angle and if you chose to work with that you will get the same solutions. We are using the positive angle only to make sure we don’t accidentally lose the minus sign on the angle we received from our calculator.
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,
\[\frac{{4z}}{9} = 2.7367 + 2\pi n\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}\frac{{4z}}{9} = 5.8783 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \]Finally, to get all the solutions to the equation all we need to do is multiply both sides by \(\frac{9}{4}\) and we’ll convert everything to decimals to help with the final step.
\[\begin{align*}z & = 6.1576 + \frac{{9\pi n}}{2} & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} & \hspace{0.15in} z = 13.2262 + \frac{{9\pi n}}{2}\hspace{0.25in} & n = 0, \pm 1, \pm 2, \ldots \\ & = 6.1576 + 14.1372n & \hspace{0.25in}{\mbox{OR }}\hspace{0.25in} & \hspace{0.25in} = 13.2262 + 14.1372n\hspace{0.25in} & n = 0, \pm 1, \pm 2, \ldots \end{align*}\]Now let’s find all the solutions.
\[\begin{array}{lclcl}{n = - 1:\,} & {z = - 7.9796\,} & \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} &{z = - 0.9110}\\{n = 0:} &{z = 6.1576\,}& \hspace{0.25in} {{\mbox{OR}}} \hspace{0.25in} & \hspace{0.35in} {\require{cancel} \xcancel{{z = 13.2262}} > 10}\end{array}\]Notice that with each increase in \(n\) we were really just adding/subtracting (depending on the sign of \(n\)) another 14.1372 onto 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 three solutions to this equation in the given interval.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{z = - 7.9796,\,\, - 0.9110,\,\,6.1576}}\]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.