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Section 1.4 : Solving Trig Equations

4. Without using a calculator find the solution(s) to \(\displaystyle 2\cos \left( {\frac{x}{3}} \right) + \sqrt 2 = 0\) that are in \(\left[ { - 7\pi ,7\pi } \right]\).

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Hint : First, find all the solutions to the equation without regard to the given interval.
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Because we found all the solutions to this equation in Problem 3 of this section we’ll just list the result here. For full details on how these solutions were obtained please see the solution to Problem 3.

All solutions to the equation are,

\[x = \frac{{9\pi }}{4} + 6\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} x = \frac{{15\pi }}{4} + 6\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]
Hint : Now all we need to do is plug in values of \(n\) to determine which solutions will actually fall in this interval.
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Note that because at least some of the solutions will have a denominator of 4 it will probably be convenient to also have the interval written in terms of fractions with denominators of 4. Doing this will make it much easier to identify solutions that fall inside the interval so,

\[\left[ { - 7\pi ,7\pi } \right] = \left[ { - \frac{{28\pi }}{4},\frac{{28\pi }}{4}} \right]\]

With the interval written in this form, if our potential solutions have a denominator of 4, all we need to do is compare numerators. As long as the numerators are between \( - 28\pi \) and \(28\pi \) we’ll know that the solution is in the interval.

Also, in order to quickly determine the solution for particular values of \(n\) it will be much easier to have both fractions in the solutions have denominators of 4. So, the solutions, written in this form, are.

\[x = \frac{{9\pi }}{4} + \frac{{24\pi n}}{4} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} x = \frac{{15\pi }}{4} + \frac{{24\pi n}}{4} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Now let’s find all the solutions.

\[\begin{array}{llcl}{n = - 2:} & {x = \require{cancel} \xcancel{{ - \frac{{39\pi }}{4}}} < - \frac{{28\pi }}{4}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {x = \xcancel{{ - \frac{{33\pi }}{4}}} < - \frac{{28\pi }}{4}}\\ {n = - 1:} & {x = - \frac{{15\pi }}{4}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {x = - \frac{{9\pi }}{4}}\\ {n = 0:} & {x = \frac{{9\pi }}{4}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {x = \frac{{15\pi }}{4}}\\ {n = 1:} & {x = \xcancel{{\frac{{33\pi }}{4}}} > \frac{{28\pi }}{4}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {x = \xcancel{{\frac{{39\pi }}{4}}} > \frac{{28\pi }}{4}}\end{array}\]

Note that we didn’t really need to plug in \(n = 1\) or \(n = - 2\) above to see that they would not work. With each increase in \(n\) we were really just adding (for positive \(n\)) or subtracting (for negative \(n\)) another \(\frac{{24\pi }}{4}\) from the previous results. By a quick inspection we could see that adding \(24\pi \) to the numerator of either solution from the \(n = 1\) step would result in a numerator that is larger than \(28\pi \) and so would result in a solution that is outside of the interval. Likewise, for the \(n = - 2\) case, subtracting \(24\pi \) from each of the numerators will result in numerators that will be less than \( - 28\pi \) and so will not be in the interval. This is not something that must be noticed in order to work the problem, but noticing this would definitely help reduce the amount of actual work.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{15\pi }}{4},\, - \frac{{9\pi }}{4},\,\,\frac{{9\pi }}{4},\,\,\frac{{15\pi }}{4}}}\]