SOLVING TRIGONOMETRIC EQUATIONS


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


Problem 9.8c:        Solve for x in the equation

\begin{displaymath}8\sin \left( 5x-1\right) =1\end{displaymath}


Answer:    The exact answers are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &=&\displaystyle \frac{1+\s...
...isplaystyle \displaystyle \frac{%
2n\pi }{5} \\
&&
\end{array}\end{displaymath}

where n is an integer.




The approximate values of these solutions are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &\approx &0.2250656\pm 1.25...
... \\
x_{2} &\approx &0.80325296\pm 1.256637n \\
&&
\end{array}\end{displaymath}

$\quad $where n is an integer.




Solution:


There are an infinite number of solutions to this problem.



First isolate the sine term.

\begin{displaymath}\begin{array}{rclll}
8\sin \left( 5x-1\right) &=&1 \\
&& \\ ...
...ft( 5x-1\right) &=&\displaystyle \frac{1}{8} \\
&&
\end{array}\end{displaymath}

To solve for x, we have to isolate x. How do we isolate the x? We could take the inverse (arcsine) of both sides. However, inverse functions can only be applied to one-to-one functions and the sine function is not one-to-one.


Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. The sine 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 sine function to that interval , we can take the arcsine of both sides of each equation.


\begin{displaymath}\begin{array}{rclll}
&& \\
-\displaystyle \displaystyle \fra...
...yle \frac{1}{8}\right) }{5}\approx 0.2250656 \\
&&
\end{array}\end{displaymath}

The angle x is the reference angle. We know that

\begin{displaymath}\begin{array}{rclll}
&& \\
\sin \left( 5x-1\right) &=&\sin \left( \pi -\left( 5x-1\right) \right) . \\
&&
\end{array}\end{displaymath}

Therefore, if $\sin \left( 5x-1\right) =\displaystyle \displaystyle \frac{1}{8}$, then $\sin \left( \pi
-\left( 5x-1\right) \right) =\displaystyle \displaystyle \frac{1}{8}.$


\begin{displaymath}\begin{array}{rclll}
&& \\
\sin \left( \pi -\left( 5x-1\righ...
...le \frac{1}{8}\right) }{5}\approx 0.80325296
\\
&&
\end{array}\end{displaymath}

The period of $\sin (x)$ equals $2\pi $ and the period of $\sin \left(
5x-1\right) $ equals $\displaystyle \displaystyle \frac{2\pi }{5}$, this means other solutions exists every $\pm \displaystyle \displaystyle \frac{2\pi }{5}$ units. The exact solutions are

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&\displaystyle \frac{1+\sin ^{-1...
...isplaystyle \displaystyle \frac{%
2n\pi }{5} \\
&&
\end{array}\end{displaymath}

where n is an integer.




The approximate values of these solutions are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &\approx &0.2250656\pm 1.25...
... \\
x_{2} &\approx &0.80325296\pm 1.256637n \\
&&
\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 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=0.2250656


Left Side:

\begin{displaymath}8\sin \left( 5x-1\right) \approx 8\sin \left( 5\left(
0.2250656\right) -1\right) \approx 1\end{displaymath}

Right Side:        $1\bigskip $

Since the left side of the original equation equals the right side of the original equation when you substitute 0.2250656 for x, then 0.2250656 is a solution.




Check solution x=0.80325296


Left Side:

\begin{displaymath}8\sin \left( 5x-1\right) \approx 8\sin \left( 5\left(
0.80325296\right) -1\right) \approx 1\end{displaymath}

Right Side:        $1\bigskip $

Since the left side of the original equation equals the right side of the original equation when you substitute 0.80325296 for x, then 0.80325296is a solution.




We have just verified algebraically that the exact solutions are $x=\displaystyle \displaystyle \frac{%
1+\sin ^{-1}\left( \displaystyle \frac{1}{8}\right) }{5}$ and $x=\displaystyle \displaystyle \frac{\pi +1-\sin
^{-1}\left( \displaystyle \displaystyle \frac{1}{8}\right) }{5}$ and these solutions repeat every $\pm \displaystyle \displaystyle \frac{2\pi }{5}$ units. The approximate values of these solutions are $%
x\approx 0.2250656$ and 0.80325296 and these solutions repeat every $\pm
1.256637$ units.




Graphical Check:


Graph the equation $f(x)=8\sin \left( 5x-1\right) -1$ (formed by subtracting the right side of the original equation from the left side of the original equation). 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 0.2250656. Since the period is $%
\displaystyle \displaystyle \frac{2\pi }{5}\approx 1.256637$, you can verify that the graph also crosses the x-axis again at $0.2250656+1.256637\approx 1.4817026$ and at $%
0.2250656+2\left( 1.256637\right) =2.7383396$, 3.9949766, 5.2516136 etc.


Verify the graph crosses the x-axis at 0.80325296. Since the period is $%
\displaystyle \displaystyle \frac{2\pi }{5}\approx 1.256637$, you can verify that the graph also crosses the x-axis again at $0.80325296+1.256637\approx 2.05988996$ and at $%
0.80325296+2\left( 1.256637\right) =3.31652696$, 4.57316396, 5.82980096etc.


Note: If the problem were to find the solutions in the interval $\left[
0,3\right] $, then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,3\right] :$ $x\approx 0.2250656,$ 0.80325296, 1.4817026, 2.05988996, and $2.7383396.\bigskip\bigskip
\bigskip\bigskip $

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Author: Nancy Marcus

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