SOLVING TRIGONOMETRIC EQUATIONS


Note:

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



Solve for x in the following equation.


Example 4:

$\sin \left( \displaystyle \frac{11}{5}x\right) -\displaystyle \frac{2}{7}=0$


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

\begin{eqnarray*}&& \\
\sin \left( \displaystyle \frac{11}{5}x\right) -\display...
...11}{5}x\right) &=&\displaystyle \frac{2}{7} \\
&& \\
&& \\
&&
\end{eqnarray*}
If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2}
\leq \displaystyle \frac{11}{5}x\leq ,\dis...
...isplaystyle \frac{
5\pi }{22}\leq x\leq ,\displaystyle \frac{5\pi }{22}\right] $, we can use the inverse sine function to solve for reference angle $\displaystyle \frac{11}{5}x$ and then x.
\begin{eqnarray*}&& \\
\sin \left( \displaystyle \frac{11}{5}x\right) &=&\displ...
...ght) &=&\sin
^{-1}\left( \displaystyle \frac{2}{7}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
\displaystyle \frac{11}{5}x &=&\sin ^{-1}\left( \displays...
...x &\approx &0.131705318835\ \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 $\displaystyle \frac{11}{5}x$ 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{11}{5}x\right) &=&\...
...{5}x &=&\sin ^{-1}\left( \displaystyle \frac{2}{7}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
\displaystyle \frac{11}{5}x &=&\pi -\sin ^{-1}\left( \dis...
...&& \\
x &\approx &1.29629134189\ \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\bigskip $

The period of the sin $\left( \displaystyle \frac{11}{5}x\right) $ function is $\displaystyle \frac{10}{
11}\pi .$ This means that the values will repeat every $\displaystyle \frac{10}{11}\pi $radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{5}{
11}\sin ^{-1}\left( \displaystyle \frac{2}{7}\right) \pm n\left( \displaystyle \frac{10}{11}\pi
\right) $ and $x=\displaystyle \frac{5}{11}\pi -\displaystyle \frac{5}{11}\sin ^{-1}\left( \displaystyle \frac{2}{7}
\right) \pm n\left( \displaystyle \frac{10}{11}\pi \right) $ where n is an integer.




The approximate solutions are $x\approx 0.131705318835\pm n\left( \displaystyle \frac{10}{
11}\pi \right) $ and $x\approx 1.29629134189\pm n\left( \displaystyle \frac{10}{11}\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=0.131705318835


Since the left side equals the right side when you substitute 0.131705318835 for x, then 0.131705318835 is a solution.




Check the answer x=1.29629134189

Since the left side equals the right side when you substitute 1.29629134189for x, then 1.29629134189 is a solution.




Graphical Check:


Graph the equation

$f(x)=\sin \left( \displaystyle \frac{11}{5}x\right) -\displaystyle \frac{2}{7}.$

Note that the graph crosses the x-axis many times indicating many solutions. Note that it crosses at 0.131705318835. Since the period is $\displaystyle \frac{10}{11}
\pi \approx 2.85599$, it crosses again at 0.131705318835 + 2.85599 = 2.98769864028 and at 0.131705318835 + 2( 2.85599 ) = 5.843685, etc.


The graph crosses at 1.29629134189. Since the period is $\displaystyle \frac{10}{11}
\pi \approx 2.85599$, it will cross again at $1.29629134189+\left(
2.85599\right) =4.152281$ and at $1.29629134189+2\left( 2.85599\right)
=7.00827$, etc.



If you would like to test yourself by working some problems similar to this example, click on Problem.


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