SOLVING TRIGONOMETRIC EQUATIONS


Note:

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



Solve for x in the following equation.


Example 1:        

$2\sin\left( 3x\right) -1=0$



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

\begin{eqnarray*}&& \\
2\sin \left( 3x\right) -1 &=&0 \\
&& \\
\sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\
&& \\
&&
\end{eqnarray*}


If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2}
\leq 3x\leq ,\displaystyle \frac{\pi }{2}\...
...[ -\displaystyle \frac{\pi }{6}\leq
x\leq ,\displaystyle \frac{\pi }{6}\right] $, we can use the inverse sine function to solve for reference angle 3x and then x.

\begin{eqnarray*}&& \\
\sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\
&...
...&=&\sin ^{-1}\left( \displaystyle \frac{1}{2
}\right) \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\
&& ...
...\
x &\approx &0.174532925\ \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 3x that terminates in the first quadrant and the angle $\pi -3x$ that terminates in the second quadrant. We have already solved for 3x.

\begin{eqnarray*}&& \\
\sin \left( \pi -3x\right) &=&\displaystyle \frac{1}{2} ...
...-3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
3x &=&\pi -\sin ^{-1}\left( \displaystyle \frac{1}{2}\ri...
... \\
&& \\
x &\approx &0.872665\ \mbox{ radians } \\
&& \\
&&
\end{eqnarray*}

The solutions are $x=\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) $ and $
x=\displaystyle \frac{\pi }{3}-\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) .\bigskip
\bigskip\bigskip $

The period of the $\sin\left( 3x\right) $ function is $\displaystyle \frac{2\pi }{3}.$This means that the values will repeat every $\displaystyle \frac{2\pi }{3}$ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{1}{3}\sin
^{-1}\left( \displaystyle \frac{1}{2}\right) \pm n\left( \displaystyle \frac{2\pi }{3}\right) $ and $x=
\displaystyle \frac{\pi }{3}-\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \pm n\left(
\displaystyle \frac{2\pi }{3}\right) $ where n is an integer.


The approximate solutions are $x=0.174532925\pm n\left( \displaystyle \frac{2\pi }{3}
\right) $ and $x\approx 0.872665\pm n\left( \displaystyle \frac{2\pi }{3}\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=0.174532925


Since the left side equals the right side when you substitute 0.174532925for x, then 0.174532925 is a solution.




Check the answer x=0.872665


Since the left side equals the right side when you substitute $x\approx
0.872665$ for x, then $x\approx
0.872665$ is a solution.



Graphical Check:


Graph the equation

$f(x)=2\sin (3x)-1.$

Note that the graph crosses the x-axis many times indicating many solutions. You can see that the graph crosses at 0.174532925. Since the period is $\displaystyle \frac{2\pi }{3}\approx
2.094395$, it crosses again at 0.174532925+2.094395=2.2689 and at 0.174532925+2(2.094395)=4.3633, etc. The graph crosses at 0.872665.


Since the period is $\displaystyle \frac{2\pi }{3}\approx
2.094395$, it will cross again at $0.872665+\left( 2.094395\right) =2.9671$ and at 0.872665+2(2.094395)=5.061455, 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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