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### Section 1.6 : Solving Trig Equations with Calculators, Part II

1. Find all the solutions to $$3 - 14\sin \left( {12t + 7} \right) = 13$$. Use at least 4 decimal places in your work.

Hint : With the exception of the argument, which is a little more complex, this is identical to the equations that we solved in the previous section.
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The argument of the sine is a little more complex in this equation than those we saw in the previous section, but the solution process is identical. Therefore, we will be assuming that you recall the process from the previous section and do not need all the hints or quite as many details as we put into the solutions there. 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 section.

First, isolating the sine on one side of the equation gives,

$\sin \left( {12t + 7} \right) = - \frac{5}{7}$

Using a calculator we get,

$12t + 7 = {\sin ^{ - 1}}\left( { - \frac{5}{7}} \right) = - 0.7956$

From our knowledge of the unit circle we can see that a positive angle that corresponds to this angle is $$2\pi - 0.7956 = 5.4876$$. Either these angles can be used here but we’ll use the positive angle to avoid the possibility of losing the minus sign. Also, from a quick look at a unit circle we can see that a second angle in the range $$\left[ {0,2\pi } \right]$$ will be $$\pi + 0.7965 = 3.9372$$.

Now, all possible angles for which sine will have this value are,

$12t + 7 = 3.9372 + 2\pi n\hspace{0.25in}{\rm{OR }}\hspace{0.25in}12t + 7 = 5.4876 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots$

At this point all we need to do is solve each of these for $$t$$ and we’ll have all the solutions to the equation. Doing this gives,

$\require{bbox} \bbox[2pt,border:1px solid black]{{t = - 0.2552 + \frac{{\pi n}}{6}\hspace{0.25in}{\rm{OR }}\hspace{0.25in}t = - 0.1260 + \frac{{\pi n}}{6}\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }}$

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.