SOLVING TRIGONOMETRIC EQUATIONS


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


Problem 9.7d:        Solve for x in the equation

\begin{displaymath}5\sec ^{2}x-11\tan x-4=0\end{displaymath}


Answer:    The exact answers are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &=&\tan ^{-1}\left( \displa...
...yle \frac{11-\sqrt{101}}{10}\right) \pm n\pi \\
&&
\end{array}\end{displaymath}

where n is an integer.




The approximate values of these solutions are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &\approx &1.127297\pm 3.141...
...&& \\
x_{2} &\approx &0.94728\pm 3.1415927n \\
&&
\end{array}\end{displaymath}

$\quad $where n is an integer.




Solution:


There are an infinite number of solutions to this problem.



In the equation $5\sec ^{2}x-11\tan x-4=0,$ there are two different trigonometric terms. To simplify the solution, convert the equation to an equivalent equation with like trigonometric terms. Which trigonometric term shall it be? Use the identity $\tan ^{2}\left( x\right) +1=\sec ^{2}\left(
x\right) $ to convert the secant term to a tangent term. Substitute $\tan
^{2}\left( x\right) +1$ for $\sec ^{2}\left( x\right) $ in the original equation.


\begin{displaymath}\begin{array}{rclll}
&& \\
5\sec ^{2}x-11\tan x-4 &=&0 \\
&...
...\
&& \\
5\tan ^{2}x-11\tan x+1 &=&0 \\
&& \\
&&
\end{array}\end{displaymath}

Let's now solve for $\tan \left( x\right) .$ The equation $5\tan
^{2}x-11\tan x+1=0$ is not easily factored, so we'll solve for $\tan \left(
x\right) $ using the Quadratic Formula.

\begin{displaymath}\begin{array}{rclll}
&& \\
5\tan ^{2}x-11\tan x+1 &=&0 \\
&...
...&=&\displaystyle \frac{11\pm \sqrt{101}}{10} \\
&&
\end{array}\end{displaymath}

To solve for x, we must first isolate the x? How do we do that? We could take the $\arctan $ of both sides. However, the tangent function is not a one-to-one function.


We could restrict the domain of the function so it is one-to-one on the restricted domain while preserving the original range. 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{displaymath}\begin{array}{rclll}
(1)\qquad \tan \left( x\right) &=&\displ...
...rt{101}}{10}\right) \approx 0.9472807 \\
&& \\
&&
\end{array}\end{displaymath}

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

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&\tan ^{-1}\left( \displaystyle ...
...yle \frac{11-\sqrt{101}}{10}\right) \pm n\pi \\
&&
\end{array}\end{displaymath}

where n is an integer.




The approximate values of these solutions are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &\approx &1.127297\pm 6.283...
...&& \\
x_{2} &\approx &0.94728\pm 6.2831853n \\
&&
\end{array}\end{displaymath}

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 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 <tex2htmlcommentmark> 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 $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{11+\sqrt{101}}{10}\right) \approx
1.127297$


Left Side:

\begin{displaymath}5\sec ^{2}x-11\tan x-4\approx 5\sec ^{2}\left(
1.127297\right) -11\tan \left( 1.127297\right) -4\approx 0\end{displaymath}

Right Side:        $0\bigskip $

Since the left side of the original equation equals the right side of the original equation when you substitute 1.127297 for x, then 1.127297 is a solution.




Check solution $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{11-\sqrt{101}}{10}\right) \approx
0.0947281$


Left Side:

\begin{displaymath}5\sec ^{2}x-11\tan x-4\approx 5\sec ^{2}\left(
0.0947281\right) -11\tan \left( 0.0947281\right) -4\approx 0\end{displaymath}

Right Side:        $0\bigskip $

Since the left side of the original equation equals the right side of the original equation when you substitute 0.0947281 for x, then 0.0947281 is a solution.




The exact solutions $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{11\pm \sqrt{101}}{10}\right)
\ $and these solutions repeat every $\pm \pi $ units. The approximate values of these solutions are $x\approx 0.094728$ and 1.127297 and these solutions repeat every $\pm 3.14159265$ units.




Graphical Check:


Graph the equation $f(x)=5\sec ^{2}x-11\tan x-4.$ Note that the graph crosses the x-axis many times indicating many solutions.


Verify the graph crosses the x-axis at 1.127297. Since the period is $\pi
\approx 3.14159265$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 1.127297+3.14159265=4.26888965 and at $1.127297+2\left( 3.14159265\right)
=7.4104823$, etc.


The graph crosses the x-axis at $\ 0.0947281$. Since the period is $\pi
\approx 3.14159265$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 0.0947281+3.14159265=3.2363208 and at $0.0947281+2\left( 3.14159265\right)
=6.3779134$, 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 $\left[ 0,2\pi \right] :$ $x\approx
0.0947281,$ 1.127297, 3.2363208 and $4.26888965.\bigskip\bigskip
\bigskip\bigskip $

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Author: Nancy Marcus

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