SOLVING TRIGONOMETRIC EQUATIONS

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

Example 4:        Solve for x in the following equation.

$$15\tan ^{4}\left( x\right) +\tan ^{3}x-36\tan ^{2}x-7\tan x+15=0$$

There are an infinite number of solutions to this problem.

Let's manipulate the equation to make solving for x a little easier. Rewrite the left side of the equation in an equivalent factored form.

$$\begin{array}{rclll} && \\ 15\tan ^{4}\left( x\right) +\tan ... ...ht) \left( 5\tan ^{2}x+2\tan x-3\right) &=&0 \\ && \end{array}$$

The product of factors equals zero if at least one of the factors equals zero. This means that

$$\begin{array}{rclll} && \\ \left( 3\tan ^{2}x-\tan x-5\right... ... &&or \\ 5\tan ^{2}x+2\tan x-3 &=&0 \\ && \\ && \end{array}$$

Solve for $\tan x$ in each of these equations.

$$\begin{array}{rclll} \left( 1\right) \qquad 3\tan ^{2}x-\tan ... ...4}+1 &=&0 \\ && \\ \tan x_{4} &=&-1 \\ && \\ && \end{array}$$

$\$

How do we isolate the x in each of these equations? We could take the inverse (arctangent) of both sides of each equation. However, the tangent function is not a one-to-one function.

Let's restrict the domain so 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 $\left( -\displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{\pi }{% 2}\right) .$ If we restrict the domain of the tangent function to that interval , we can take the arctangent of both sides of each equation.

$$\begin{array}{rclll} \tan x_{1} &=&\displaystyle \frac{1+\sqr... ...-1}\left( -1\right) \approx -0.785398 \\ && \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ x_{1} &=&\tan ^{-1}\left( \displa... ...x_{4} &=&\tan ^{-1}\left( -1\right) \pm n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.972919\pm 3.141... ...& \\ x_{4} &\approx &-0.785398\pm 3.141593n \\ && \end{array}$$

where n is an integer.

You 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.

You can also check the solutions graphically by graphing the function formed by subtracting the right side of the original equation from the left side of the original equation. The solutions of the original equation are the x-intercepts of this graph.

Algebraic Check:

Check solution $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{1+\sqrt{61}}{6}\right) \approx 0.972919$

Left Side:

$$\begin{array}{rclll} 15\tan ^{4}\left( x\right) +\tan ^{3}x-3... ...&&-7\tan \left( 0.972919\right) +15 \\ &\approx &0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.972919 for x, then 0.972919 is a solution.

Check solution $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{1-\sqrt{61}}{6}\right) \approx -0.848564$

Left Side:

$$\begin{array}{rclll} 15\tan ^{4}\left( x\right) +\tan ^{3}x-3... ... -7\tan \left( -0.848564\right) +15 \\ &\approx &0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -0.848564 for x, then -0.848564 is a solution.

Check solution $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{3}{5}\right) \approx 0.5404195$

Left Side:

$$\begin{array}{rclll} 15\tan ^{4}\left( x\right) +\tan ^{3}x-3... ... -7\tan \left( 0.5404195\right) +15 \\ &\approx &0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.5404195 for x, then 0.5404195 is a solution.

Check solution $x=\tan ^{-1}\left( -1\right) \approx -0.785398$

Left Side:

$$\begin{array}{rclll} 15\tan ^{4}\left( x\right) +\tan ^{3}x-3... ...&-7\tan \left( -0.785398\right) +15 \\ &\approx &0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -0.785398 for x, then -0.785398 is a solution.

We have just verified algebraically that the exact solutions are $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{1\pm \sqrt{61}}{6}\right)$, $\tan ^{-1}\left( \displaystyle \displaystyle \frac{3}{5% }\right)$and $x=\tan ^{-1}\left( -1\right) ,$ and these solutions repeat every $\pm \pi$ units. The approximate values of these solutions are $% x\approx 0.972919$, -0.848564, 0.5404195 and -0.785398 and these solutions repeat every $\pm 3.141593$ units.

Graphical Check:

Graph the function $f(x)=15\tan ^{4}\left( x\right) +\tan ^{3}x-36\tan ^{2}x-7\tan x+15$,formed by the left side of the equation. The x-intercepts of the graph are the real solutions.

Note that the graph crosses the x-axis many times indicating many solutions. Let's check a few of these x-intercepts against the solutions we derived.

Verify the graph crosses the x-axis at 0.972919. Since the period is $\pi \approx 3.141593$, you can verify that the graph also crosses the x-axis again at 0.972919+3.141593=4.114512 and at $0.972919+2\left( 3.141593\right) =7.2561$, etc.

Verify the graph crosses the x-axis at -0.848564. Since the period is $\pi \approx 3.141593$, you can verify that the graph also crosses the x-axis again at -0.848564+3.141593=2.293028 and at $-0.848564+2\left( 3.141593\right) =5.43462$, etc.

Verify the graph crosses the x-axis at 0.5404195. Since the period is $\pi \approx 3.141593$, you can verify that the graph also crosses the x-axis again at 0.5404195+3.141593=3.68201 and at $0.5404195+2\left( 3.141593\right) =6.8236$, etc.

Verify the graph crosses the x-axis at -0.785398. Since the period is $\pi \approx 3.141593$, you can verify that the graph also crosses the x-axis again at -0.785398+3.141593=2.356195 and at $-0.785398+2\left( 3.141593\right) =5.497787$, etc.

Note: If the problem were to find the solutions in the interval $\left[ 0,\pi \right]$, then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,\pi \right] :$ $x\approx 0.5404195,$ 0.972919, 2.293028 and $2.356195.\bigskip\bigskip\bigskip \bigskip$

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

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