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### Section 1.4 : Solving Trig Equations

2. Without using a calculator find the solution(s) to $$4\sin \left( {3t} \right) = 2$$ that are in $$\displaystyle \left[ {0,\frac{{4\pi }}{3}} \right]$$.

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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 1 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 1.

All solutions to the equation are,

$t = \frac{\pi }{{18}} + \frac{{2\pi n}}{3} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} t = \frac{{5\pi }}{{18}} + \frac{{2\pi n}}{3} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots$
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 18 it will probably be convenient to also have the interval written in terms of fractions with denominators of 18. Doing this will make it much easier to identify solutions that fall inside the interval so,

$\left[ {0,\frac{{4\pi }}{3}} \right] = \left[ {0,\frac{{24\pi }}{{18}}} \right]$

With the interval written in this form, if our potential solutions have a denominator of 18, all we need to do is compare numerators. As long as the numerators are positive and less than $$24\pi$$ 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 18. So, the solutions, written in this form, are.

$t = \frac{\pi }{{18}} + \frac{{12\pi n}}{{18}}\hspace{0.5in}{\rm{OR }}\hspace{0.5in} t = \frac{{5\pi }}{{18}} + \frac{{12\pi n}}{{18}} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots$

Now let’s find all the solutions. First notice that, in this case, if we plug in negative values of $$n$$ we will get negative solutions and these will not be in the interval and so there is no reason to even try these. So, let’s start at $$n = 0$$ and see what we get.

$\require{cancel} \begin{array}{llcl}{n = 0:} & {t = \frac{\pi }{{18}}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {t = \frac{{5\pi }}{{18}}}\\ {n = 1:} & {t = \frac{{13\pi }}{{18}}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {t = \frac{{17\pi }}{{18}}}\\ {n = 2:} & {t = \xcancel{{\frac{{25\pi }}{{18}}}} > \frac{{24\pi }}{{18}}\,} & {\,\,\,{\rm{OR}}\,\,\,} & {t = \xcancel{{\frac{{29\pi }}{{18}}}} > \frac{{24\pi }}{{18}}}\end{array}$

Note that we didn’t really need to plug in $$n = 2$$ above to see that they would not work. With each increase in $$n$$ we were really just adding another $$\frac{{12\pi }}{{18}}$$ onto the previous results and by a quick inspection we could see that adding $$12\pi$$ to the numerator of either solution from the $$n = 1$$ step would result in a numerator that is larger than $$24\pi$$ and so would result in a solution that is outside of 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.

$\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{\pi }{{18}},\,\,\frac{{5\pi }}{{18}},\,\,\frac{{13\pi }}{{18}},\,\,\frac{{17\pi }}{{18}}}}$