SOLVING TRIGONOMETRIC EQUATIONS


Note:

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



Solve for the real number x in the following equation.



Problem 9.3b:         $17\cos \left( 6x\right) -8=7\cos \left( 6x\right) $


Answers:        There are an infinite number of solutions: $x=\displaystyle \frac{1}{6}
\cos ^{-1}\left( \displaystyle \frac{4}{5}\right) \pm n\left( \displaystyle \frac{\pi }{3}\right) $and $x=\displaystyle \frac{\pi }{3}-\cos ^{-1}\left( \displaystyle \frac{4}{5}\right) \pm n\left(
\displaystyle \frac{\pi }{3}\right) $ are the exact solutions, and $x\approx 0.10725\pm
n\left( \displaystyle \frac{\pi }{3}\right) $ and $x\approx 0.9399473664\pm n\left(
\displaystyle \frac{\pi }{3}\right) $ are the approximate solutions.



Solution:


To solve for x, first isolate the cosine term.


\begin{eqnarray*}&& \\
17\cos \left( 6x\right) -8 &=&7\cos \left( 6x\right) \\ ...
...isplaystyle \frac{8}{10}=\displaystyle \frac{4}{5} \\
&& \\
&&
\end{eqnarray*}


If we restrict the domain of the cosine function to $0\leq 6x\leq \pi
\rightarrow 0\leq x\leq \displaystyle \frac{\pi }{6}$, we can use the arccos function to solve for x.

\begin{eqnarray*}\cos \left( 6x\right) &=&\displaystyle \frac{8}{10}=\displaysty...
...ac{1}{6}\cos ^{-1}\left( \displaystyle \frac{4}{5}\right) \\
&&
\end{eqnarray*}



\begin{eqnarray*}&&\\
\mbox{ Reference Angle } &:&x^{\prime }=\displaystyle \fr...
...\
\mbox{ Reference Angle } &:&x^{\prime }=0.10725 \\
&& \\
&&
\end{eqnarray*}


The period of $\cos \left( x\right) $ is $2\pi .$ The period of $\cos \left(
6x\right) $ is $\displaystyle \frac{2\pi }{6}=\displaystyle \frac{\pi }{3}.$ Divide the interval $
\left[ 0,\displaystyle \frac{\pi }{3}\right] $ into four equal intervals: $\left[ 0,
\displaystyle \frac{\pi }{12}\right] ,\ \left[ \displaystyle \frac{\p...
... ,\
\left[ \displaystyle \frac{\pi }{6},\displaystyle \frac{\pi }{4}\right] ,\ $and $\left[ \displaystyle \frac{\pi }{4}
,\displaystyle \frac{\pi }{3}\right] .\bigskip\bigskip $

The cosine of 6x is positive in the first quadrant $\left[ 0,\displaystyle \frac{\pi }{12}
\right] $ and in the fourth quadrant $\left[ \displaystyle \frac{\pi }{4},\displaystyle \frac{\pi }{3}
\right] $.



This means that there are two solutions in the first counterclockwise rotation from 0 to $\displaystyle \frac{\pi }{3}$. One angle, $x_{1}=x^{\prime }=\cos
^{-1}\left( \displaystyle \frac{4}{5}\right) \approx 0.10725$ terminates in the first quadrant and angle $x_{2}=\displaystyle \frac{\pi }{3}-\cos ^{-1}\left( \displaystyle \frac{4}{5}
\right) \approx 0.9399473664$ terminates in the fourth quadrant.


Since the period is $\displaystyle \frac{\pi }{3},$ this means that the values will repeat every $\displaystyle \frac{\pi }{3}$ radians. Therefore, the solutions are $x\approx 0.10725\pm
n\left( \displaystyle \frac{\pi }{3}\right) $ and $x\approx
0.9399474\pm n\left( \displaystyle \frac{\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=0.10725


Left Side: $\qquad 17\cos \left( 6x\right) -8\approx 17\cos \left( 6\left(
0.10725\right) \right) -8\approx 5.6000$

Right Side:         $7\cos \left( 6x\right) \approx 7\cos \left( 6\left(
0.10725\right) \right) \approx 5.6000\bigskip $

Since the left side equals the right side when you substitute 0.10725 for x, then 0.10725 is a solution.




Check the answer . x=0.9399474


Left Side: $\qquad 17\cos \left( 6x\right) -8\approx 17\cos \left( 6\left(
0.9399474\right) \right) -8\approx 5.6000$

Right Side:         $7\cos \left( 6x\right) \approx 7\cos \left( 6\left(
0.9399474\right) \right) \approx 5.6000\bigskip $

Since the left side equals the right side when you substitute 0.9399474for x, then 0.9399474 is a solution.




Graphical Check: Graph the equation $f(x)=10\cos \left( 6x\right) -8.$ (Formed by subtracting the right side of the original equation from the left side of the original equation.


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


Note the graph crosses at 0.10725 (one of the solutions). Since the period of the function is $\displaystyle \frac{\pi }{3}\approx
1.04719755$, the graph crosses again at 0.10725+1.04719755=1.15444755 and again at $0.10725+2\left( 1.04719755\right) \approx 2.2016451$, etc.


Note the graph also crosses at 0.9399474 (one of the solutions). Since the period of the function is $\displaystyle \frac{\pi }{3}\approx
1.04719755$, the graph crosses again at 0.9399474+1.04719755=1.987145 and again at $0.9399474+2\left( 1.04719755\right) \approx 3.0343425$, etc.


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




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Author: Nancy Marcus

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