SOLVING TRIGONOMETRIC EQUATIONS


Note:

If you would like an review of trigonometry, click on trigonometry.



Solve for x in the following equation.


Example 2:        

$7\sin \left( 5x\right) -1=0$


There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.

\begin{eqnarray*}&& \\
7\sin \left( 5x\right) -1 &=&0 \\
&& \\
\sin \left( 5x\right) &=&\displaystyle \frac{1}{7} \\
&& \\
&&
\end{eqnarray*}

If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2}
\leq 5x\leq ,\displaystyle \frac{\pi }{2}\...
...-\displaystyle \frac{\pi }{10}\leq
x\leq ,\displaystyle \frac{\pi }{10}\right] $, we can use the inverse sine function to solve for reference angle 5x and then x.

\begin{eqnarray*}&& \\
\sin \left( 5x\right) &=&\displaystyle \frac{1}{7} \\
&...
...ht) &=&\sin ^{-1}\left( \displaystyle \frac{1}{7
}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
5x &=&\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \...
...x &\approx &0.028669513781\ \mbox{ radians }\\
&& \\
&& \\
&&
\end{eqnarray*}

We know that the $\sin $e function is positive in the first and the second quadrant. Therefore two of the solutions are the angle 5x that terminates in the first quadrant and the angle $\pi -5x$ that terminates in the second quadrant. We have already solved for 5x.

\begin{eqnarray*}&& \\
\sin \left( \pi -5x\right) &=&\displaystyle \frac{1}{7} ...
... -5x &=&\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
5x &=&\pi -\sin ^{-1}\left( \displaystyle \frac{1}{7}\rig...
...& \\
x &\approx &0.599649016937\ \mbox{ radians } \\
&& \\
&&
\end{eqnarray*}

The exact solutions are $x=\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) $and $x=\displaystyle \frac{\pi }{5}-\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right)
.\bigskip\bigskip\bigskip $

The period of the sin $\left( 5x\right) $ function is $\displaystyle \frac{2\pi }{5}.$This means that the values will repeat every $\displaystyle \frac{2\pi }{5}$ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{1}{5}\sin
^{-1}\left( \displaystyle \frac{1}{7}\right) \pm n\left( \displaystyle \frac{2\pi }{5}\right) $ and $x=
\displaystyle \frac{\pi }{5}-\displaystyle \frac{1}{5}\sin ^{-1}\left( \displaystyle \frac{1}{7}\right) \pm n\left(
\displaystyle \frac{2\pi }{5}\right) $ where n is an integer.


The approximate solutions are $x\approx 0.0286695137811\pm n\left( \displaystyle \frac{
2\pi }{5}\right) $ and $x\approx 0.599649016937\pm n\left( \displaystyle \frac{2\pi }{5}
\right) $ where n is an integer.



These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.



Numerical Check:


Check the answer x=0.028669513781


Since the left side equals the right side when you substitute 0.028669513781 for x, then 0.028669513781 is a solution.




Check the answer x=0.599649016937


Since the left side equals the right side when you substitute $x\approx
0.599649016937$ for x, then $x\approx
0.599649016937$ is a solution.



Graphical Check:


Graph the equation

$f(x)=7\sin \left( 5x\right) -1.$

Note that the graph crosses the x-axis many times indicating many solutions.


Note that it crosses at 0.028669513781. Since the period is $\displaystyle \frac{2\pi }{
5}\approx 1.256637$, it crosses again at 0.028669513781+1.256637=1.285306and at 0.028669513781+2(1.256637)=2.54194, etc.


The graph crosses at 0.599649016937. Since the period is $\displaystyle \frac{2\pi }{
5}\approx 1.256637$, it will cross again at $0.599649016937+\left(
1.256637\right) =1.856286$ and at 0.599649016937+2(1.256637)=3.112923, etc $
.\bigskip\bigskip $



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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