SOLVING TRIGONOMETRIC EQUATIONS


Note:

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



Solve for x in the following equation.


Example 3:

$12\cos \left( 4x\right) -7=3\cos \left( 4x\right) $



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

\begin{eqnarray*}&& \\
12\cos \left( 4x\right) -7 &=&3\cos \left( 4x\right) \\ ...
...eft( 4x\right) &=&\displaystyle \frac{7}{9} \\
&& \\
&& \\
&&
\end{eqnarray*}


If we restriction the domain of the cosine function to $\left[ 0\leq 4x\leq
\pi \right] \rightarrow \left[ 0,\displaystyle \frac{\pi }{4}\right] $, we can use the inverse cosine function to solve for reference angle $x^{\prime },$ and then x.

\begin{eqnarray*}&& \\
\cos \left( 4x\right) &=&\displaystyle \frac{7}{9} \\
&...
...ht) &=&\cos ^{-1}\left( \displaystyle \frac{7}{
9}\right) \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
4x &=&\cos ^{-1}\left( \displaystyle \frac{7}{9}\right) \...
...tyle \frac{7}{9}\right) \\
&& \\
x &\approx &0.16991845 \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
\mbox{ Reference Angle } &:&x^{\prime }\approx \displayst...
...ce Angle } &:&x^{\prime }\approx 0.16991845 \\
&& \\
&& \\
&&
\end{eqnarray*}


The period of $\cos \left( x\right) $ is $2\pi $ and the period of $\cos
\left( 4x\right) $ is $\displaystyle \frac{2\pi }{4}=\displaystyle \frac{\pi }{2}.$ Divide the interval $\left[ 0,\displaystyle \frac{\pi }{2}\right] $ into four equal intervals: $
\left[ 0,\displaystyle \frac{\pi }{8}\right] ,$ $\left[ \displaystyle \frac{\pi }{8},\displaystyle \frac{\pi }{4}
\right] ,$ $\left[ \displaystyle \frac{\pi }{4},\displaystyle \frac{3\pi }{8}\right] ,$ $\left[ \displaystyle \frac{
3\pi }{8},\displaystyle \frac{\pi }{2}\right] .\bigskip $

We know that the cosine function is positive in the first and the fourth quadrant (intervals). Therefore two of the solutions are the angle $
x_{1}=x^{\prime }$ that terminates in the first quadrant and the angle $
x_{2}=\displaystyle \frac{\pi }{2}-x^{\prime }$ that terminates in the fourth quadrant. We have already solved for $x^{\prime }.$

\begin{eqnarray*}&& \\
x_{1} &=&x^{\prime }=\displaystyle \frac{1}{4}\cos ^{-1}...
...right) \\
&& \\
x_{2} &\approx &1.4008779 \\
&& \\
&& \\
&&
\end{eqnarray*}

The solutions are $x=\displaystyle \frac{1}{4}\cos ^{-1}\left( \displaystyle \frac{7}{9}\right) $ and $x=
\displaystyle \frac{\pi }{2}-\displaystyle \frac{1}{4}\cos ^{-1}\left( \displaystyle \frac{7}{9}\right) .\bigskip
\bigskip\bigskip $

The period of the $\cos
\left( 4x\right) $ function is $\displaystyle \frac{\pi }{2}.$This means that the values will repeat every $\displaystyle \frac{\pi }{2}$ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{1}{4}\cos
^{-1}\left( \displaystyle \frac{7}{9}\right) \pm n\left( 2\pi \right) $ and $x=\displaystyle \frac{\pi }{
2}-\displaystyle \frac{1}{4}\cos ^{-1}\left( \displaystyle \frac{7}{9}\right) \pm n\left( 2\pi \right) $where n is an integer.


The approximate solutions are $x\approx 0.16991845\pm n\left( 2\pi \right) $and $x\approx 1.400878\pm \left( 2\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 answer x=0.16991845


Since the left side equals the right side when you substitute 0.16991845for x, then 0.16991845 is a solution.




Check answer x=1.400878


  • Left Side: $\qquad 12\cos \left( 4x\right) -7\approx 12\cos \left( 4\left(
1.400878\right) \right) -7\approx 2.33333\bigskip $

  • Right Side:         $3\cos \left( 4x\right) \approx 3\cos \left( 4\left(
1.400878\right) \right) \approx 2.3333333\bigskip $

    Since the left side equals the right side when you substitute 1.400878 for x, then 1.400878 is a solution.



    Graphical Check:


    Graph the equation

    $f(x)=12\cos \left( 4x\right) -7-3\cos \left( 4x\right)
=9\cos \left( 4x\right) -7.$

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


    The graph crosses the x-axis at 0.16991845. Since the period is $\displaystyle \frac{
\pi }{2}\approx 1.570796$, it crosses again at 0.16991845+1.570796=1.74071477 and at 0.16991845+2(1.570796)=3.31151045, etc.


    The graph also crosses the x-axis at 1.400878. Since the period is $\displaystyle \frac{
\pi }{2}\approx 1.570796$, it crosses again at 1.400878+1.570796=2.971674and at 1.400878+2(1.570796)=4.5424707, 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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