SOLVING TRIGONOMETRIC EQUATIONS


Note:

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



Solve for x in the following equation.


Example 3:        

$8\sin \left( \displaystyle \frac{1}{5}x\right) -3=0$


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

\begin{eqnarray*}&& \\
8\sin \left( \displaystyle \frac{1}{5}x\right) -3 &=&0 \...
...e \frac{1}{5}x\right) &=&\displaystyle \frac{3}{8} \\
&& \\
&&
\end{eqnarray*}


If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2}
\leq \displaystyle \frac{1}{5}x\leq ,\disp...
...\displaystyle \frac{
5\pi }{2}\leq x\leq ,\displaystyle \frac{5\pi }{2}\right] $, we can use the inverse sine function to solve for reference angle $\displaystyle \frac{1}{5}x$ and then x.

\begin{eqnarray*}&& \\
\sin \left( \displaystyle \frac{1}{5}x\right) &=&\displa...
... &=&\sin
^{-1}\left( \displaystyle \frac{3}{8}\right) \\
&& \\
\end{eqnarray*}
\begin{eqnarray*}\displaystyle \frac{1}{5}x &=&\sin ^{-1}\left( \displaystyle \f...
...
x &\approx &1.92198387248\ \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 $\displaystyle \frac{1}{5}x$ that terminates in the first quadrant and the angle $\pi -\displaystyle \frac{1}{5}x$ that terminates in the second quadrant. We have already solved for $\displaystyle \frac{1}{5}
x. $

\begin{eqnarray*}&& \\
\sin \left( \pi -\displaystyle \frac{1}{5}x\right) &=&\d...
...{5}x &=&\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
\displaystyle \frac{1}{5}x &=&\pi -\sin ^{-1}\left( \disp...
...&& \\
x &\approx &13.7859793955\ \mbox{ radians } \\
&& \\
&&
\end{eqnarray*}

The exact solutions are $x=5\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) $ and $
x=5\pi -5\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) .\bigskip\bigskip $

The period of the sin $\left( \displaystyle \frac{1}{5}x\right) $ function is $10\pi .$This means that the values will repeat every $10\pi $ radians in both directions. Therefore, the exact solutions are $x=5\sin ^{-1}\left( \displaystyle \frac{3
}{8}\right) \pm n\left( 10\pi \right) $ and $x=5\pi -5\sin ^{-1}\left( \displaystyle \frac{
3}{8}\right) \pm n\left( 10\pi \right) $ where n is an integer.


The approximate solutions are $x\approx 1.92198387248\pm n\left( 10\pi
\right) $ and $x\approx 13.7859793955\pm n\left( 10\pi \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=1.92198387248


Since the left side equals the right side when you substitute 1.92198387248for x, then 1.92198387248 is a solution.




Check the answer x=13.7859793955


Since the left side equals the right side when you substitute 13.7859793955for x, then 13.7859793955 is a solution.



Graphical Check:


Graph the equation

$f(x)=8\sin \left( \displaystyle \frac{1}{5}x\right) -3.\bigskip $

Note that the graph crosses the x-axis many times indicating many solutions. Note that it crosses at 1.92198387248.


Since the period is \begin{displaymath}10\pi \approx 31.415927
\end{displaymath} it crosses again at 1.92198387248 + 31.415927 = 33.3379104084 and at 1.92198387248 + 2( 31.415927 ) = 64.753838 , etc. The graph crosses at 13.7859793955 . Since the period is \begin{displaymath}10\pi \approx 31.415927
\end{displaymath} , it will cross again at 13.7859793955 + ( 31.415927 ) = 45.201906 and at 13.7859793955 + 2( 31.415927) = 76.617833, etc



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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