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

6. Find all the solutions to $$\displaystyle {\tan ^2}\left( {\frac{w}{4}} \right) = \tan \left( {\frac{w}{4}} \right) + 12$$. Use at least 4 decimal places in your work.

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Hint : Factor the equation and using basic algebraic properties get two equations that can be dealt with using known techniques. If you’re not sure how to factor this think about how you would factor $${x^2} - x - 12 = 0$$.
Start Solution

This equation may look very different from anything that we’ve ever been asked to factor, however it is something that we can factor. First think about factoring the following,

${x^2} = x + 12\hspace{0.25in} \to \hspace{0.25in} {x^2} - x - 12 = \left( {x - 4} \right)\left( {x + 3} \right) = 0$

If we can factor this algebraic equation then we can factor the given equation in exactly the same manner.

\begin{align*}{\tan ^2}\left( {\frac{w}{4}} \right) & = \tan \left( {\frac{w}{4}} \right) + 12\\ {\tan ^2}\left( {\frac{w}{4}} \right) - \tan \left( {\frac{w}{4}} \right) - 12 & = 0\\ \left( {\tan \left( {\frac{w}{4}} \right) - 4} \right)\left( {\tan \left( {\frac{w}{4}} \right) + 3} \right) & = 0\end{align*}

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

$\tan \left( {\frac{w}{4}} \right) - 4 = 0\hspace{0.25in}{\rm{OR}}\hspace{0.25in}\tan \left( {\frac{w}{4}} \right) + 3 = 0$
Hint : Solve each of these two equations to attain all the solutions to the original equation.
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Each of these equations are similar to equations solved in the previous 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 the first equation and isolate the tangent to get,

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

Using our calculator we get,

$\frac{w}{4} = {\tan ^{ - 1}}\left( 4 \right) = 1.3258$

From our knowledge on solving equations involving tangents we know that the second angle in the range $$\left[ {0,2\pi } \right]$$ will be $$\pi + 1.3258 = 4.4674$$.

All the solutions to the first equation are then,

\begin{align*}\frac{w}{4} & = 1.3258 + 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in}\frac{w}{4} & = 4.4674 + 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \\w & = 5.3032 + 8\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in}w & = 17.8696 + 8\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \end{align*}

Now, let’s solve the second equation.

$\tan \left( {\frac{w}{4}} \right) = - 3\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\frac{w}{4} = {\tan ^{ - 1}}\left( { - 3} \right) = - 1.2490$

From our knowledge of the unit circle we can see that a positive angle that corresponds to this angle is $$2\pi - 1.2490 = 5.0342$$. Either these angles can be used here but we’ll use the positive angle to avoid the possibility of losing the minus sign. Also, the second angle in the range $$\left[ {0,2\pi } \right]$$ is $$\pi + \left( { - 1.2490} \right) = 1.8926$$.

All the solutions to the second equation are then,

\begin{align*}\frac{w}{4} & = 1.8926 + 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in}\frac{w}{4} & = 5.0342 + 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \\w & = 7.5704 + 8\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in}w & = 20.1368 + 8\pi n & \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]{{\begin{align*}w & = 5.3032 + 8\pi n,\hspace{0.25in} & w & = 7.5704 + 8\pi n\\w & = 17.8696 + 8\pi n,\hspace{0.25in} & w & = 20.1368 + 8\pi n\end{align*} \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.