SOLVING TRIGONOMETRIC EQUATIONS

Note: If you would like a review of trigonometry, click on trigonometry.

Example 3:        Solve for x in the following equation.

$$\tan ^{2}x-4\tan x-12=0$$

There are an infinite number of solutions to this problem.

Isolate the tangent term. To do this rewrite the left side of the equation in an equivalent factored form.

$$\begin{array}{rclll} \tan ^{2}x-4\tan x-12 &=&0 \\ && \\ \left( \tan x-6\right) \left( \tan x+2\right) &=&0 \\ \end{array}$$

The product of two factors equals zero if at least one of the factors equals zeros. This means that $\tan ^{2}x-4\tan x-12=0\$ if $\tan x-6=0$ or $\tan x+2=0.$

We have transformed a difficult problem into two problems. To find the solutions to the original equation, $\left( \tan x-6\right) \left( \tan x+2\right) =0$, we find the solutions to the equations $\tan x-6=0$ and $\tan x+2=0.$

$$\begin{array}{rclll} \tan x &=&0-6 \\ && \\ \tan x &=&6 \\ \end{array}$$

and

$$\begin{array}{rclll} \tan x+2 &=&0 \\ && \\ \tan x &=&-2 \\ \end{array}$$

How do we isolate the x? We could take the arctangent of both sides. However, the tangent function is not a one-to-one function.

Let's restrict the domain so that the function is one-to-one on the restricted domain while preserving the original range. The graph of the tangent function is one-to-one on the interval $\left( -\displaystyle \frac{\pi}{2} ,\displaystyle \frac{\pi}{2} \right)$. If we restrict the domain of the tangent function to that interval , we can take the arctangent of both sides of each equation.

$$\begin{array}{rclll} (1)\qquad \tan x &=&6 \\ && \\ \tan ^{... ...&\tan ^{-1}\left( -2\right) \approx -1.10714878 \\ \end{array}$$

Since the period of tan x equals $\pi$, these solutions will repeat every $% \pi$ units. The exact solutions are

$$\begin{array}{rclll} x_{1} &=&\tan ^{-1}\left( 6\right) \pm n... ...x_{2} &=&\tan ^{-1}\left( -2\right) \pm n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} x_{1} &\approx &1.40564765\pm 3.14159265... ...\ x_{2} &\approx &-1.1071487\pm 3.14159265n \\ && \end{array}$$

where n is an integer.

One can check each solution algebraically by substituting each solution in the original equation. If, after the substitution, the left side of the original equation equals the right side of the original equation, the solution is valid.

One can also check the solutions graphically by graphing the function formed by the left side of the original equation and graphing the function formed by the right side of the original equation. The x-coordinates of the points of intersection are the solutions. The right side of the equation is 0 and <tex2html_comment_mark> f(x)=0 is the x-axis. So really what you are looking for are the x-intercepts to the function formed by the left side of the equation.

Algebraic Check:

Check solution $x=\tan ^{-1}\left( 6\right) \approx 1.40564765$

Left Side:

$$\tan ^{2}x-4\tan x-12\approx \tan ^{2}\left( 1.40564765\right) -4\tan \left( 1.40564765\right) -12=0\approx 0$$

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute 1.40564765 for x, then 1.40564765 is a solution.

Check solution $x=\tan ^{-1}\left( -2\right) \approx -1.10714878$

Left Side:

$$\tan ^{2}x-4\tan x-12\approx \tan ^{2}\left( -1.10714878\right) -4\tan \left( -1.10714878\right) -12=0\approx 0$$

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute -1.10714878 for x, then -1.10714878 is a solution.

We have just verified that $x=\tan ^{-1}\left( 6\right) ,$ and $x=\tan ^{-1}\left( -2\right)$ are the exact solutions and these solutions repeat every $\pm \pi$ units. The approximate values of these solutions are $% x\approx 1.40564765$ and $x\approx -1.10714878$ and these solutions repeat every $\pm 3.14159265$ units.

Graphical Check:

Graph the equation $f(x)=\tan ^{2}x-4\tan x-12.$ Note that the graph crosses the x-axis many times indicating many solutions.

The graph crosses the x-axis at 1.04564765. Since the period is $\pi \approx 3.14159265$, you can verify that the graph also crosses the x-axis again at 1.04564765+3.14159265=4.54724 and at $1.04564765+2\left( 3.14159265\right) =7.688833$, etc.

The graph crosses the x-axis at -1.10714878. Since the period is $\pi \approx 3.14159265$, you can verify that the graph also crosses the x-axis again at -1.10714878+3.14159265=2.0344439 and at $-1.10714878+2\left(3.14159265\right) =5.17623659$, etc.

Note: If the problem were to find the solutions in the interval $\left[ 0,2\pi \right]$, then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,2\pi \right]$ $x\approx 1.40564765,\quad 2.034439,\quad 4.5472403,\quad and\quad 5.1760366.$

If you would like to work another example, click on Example.

If you would like to test yourself by working some problems similar to this example, click on Problem.

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