Calculus I - Trig Equations with Calculators, Part II

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Section 1.6 : Solving Trig Equations with Calculators, Part II

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3. Find all the solutions to $4\sin \left( {x + 2} \right) - 15\sin \left( {x + 2} \right)\tan \left( {4x} \right) = 0$ . Use at least 4 decimal places in your work.

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Notice that each term has a sine in it and so we can factor this out of each term to get,

$\sin \left( {x + 2} \right)\left( {4 - 15\tan \left( {4x} \right)} \right) = 0$

Now, we have a product of two factors that equals zero and so by basic algebraic properties we know that we must have,

$\sin \left( {x + 2} \right) = 0\hspace{0.25in}{\rm{OR}}\hspace{0.25in}4 - 15\tan \left( {4x} \right) = 0$ Show Step 2

Each of these equations are similar to equations solved in the previous section or in the earlier problems of this section. Therefore, we will be assuming that you can recall the solution process for each and we will not be putting in as many details. 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 solution to this problem.

We’ll start with,

$\sin \left( {x + 2} \right) = 0$

From a unit circle we can see that we must have,

$x + 2 = 0 + 2\pi n\hspace{0.25in}{\rm{OR}}\hspace{0.25in}x + 2 = \pi + 2\pi n\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots$

Notice that we can further reduce this down to,

$x + 2 = \pi n\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots$

Finally, the solutions from this equation are,

$x = \pi n - 2\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots$

The second equation will take a little more (but not much more) work. First, isolating the tangent gives,

$\tan \left( {4x} \right) = \frac{4}{{15}}$

Using our calculator we get,

$4x = {\tan ^{ - 1}}\left( {\frac{4}{{15}}} \right) = 0.2606$

From our knowledge on solving equations involving tangents we know that the second angle in the range $\left[ {0,2\pi } \right]$ will be $\pi + 0.2606 = 3.4022$ .

Finally, the solutions to this equation are,

$\begin{align*}4x & = 0.2606 + 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in} 4x & = 3.4022 + 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \\ x & = 0.06515 + \frac{{\pi n}}{2} & \hspace{0.25in}{\rm{OR}}\hspace{0.25in} x & = 0.8506 + \frac{{\pi n}}{2} & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \end{align*}$

Putting all of this together gives the following set of solutions.

$\require{bbox} \bbox[2pt,border:1px solid black]{{x = \pi n - 2, \hspace{0.25in} x = 0.06515 + \frac{{\pi n}}{2}, \hspace{0.25in} {\rm{OR }} \hspace{0.25in} x = 0.8506 + \frac{{\pi n}}{2}\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }}$

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 4^th decimal place or so however.