SOLVING TRIGONOMETRIC EQUATIONS

Note:

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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 $$\frac{2\pi }{6}=\displaystyle \frac{\pi }{3}.$$ 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 $$\frac{\pi }{3}$$. 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 $$\frac{\pi }{3},$$ this means that the values will repeat every $$\frac{\pi }{3}$$ 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 $$\frac{\pi }{3}\approx 1.04719755$$, 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 $$\frac{\pi }{3}\approx 1.04719755$$, 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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