SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.4c:        Solve for x in the following equation.

$$\displaystyle \frac{3}{5}\cos ^{2}\left( 3x\right) -\displaystyle \displaystyle \frac{1}{8}=0$$

Answers:

There are an infinite number of solutions. The exact solutions are $x=\displaystyle \displaystyle \frac{% 1}{3} \cos ^{-1}\sqrt{\displaystyle \disp... ...le \frac{5}{24}}\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}\right)$, $x=% \displaystyle \displaystyle \frac{\pi }{3}-\displaystyle \displaystyle \fra... ...\frac{5}{24}}\pm n\left( \displaystyle \displaystyle \frac{% 2\pi }{3}\right) ,$ $x=\displaystyle \displaystyle \frac{\pi }{3}+\displaystyle \displaystyle \frac{... ...\frac{5}{% 24}}\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}\right) ,$ and $x=\displaystyle \displaystyle \frac{2\pi }{3}-\displaystyle \displaystyle \frac... ...e \frac{5}{24}}\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}\right) .$

The approximate solutions are $x=0.3656038\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}% \right)$, $x=0.6815937\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}\right) ,$ $x=1.412801\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}\right) ,$ and $x=\displaystyle \displaystyle \frac{2\pi }{3}-1.72879\pm n\left( \displaystyle \displaystyle \frac{2\pi }{3}\right) .$

Solution:

To solve for x, first isolate the cosine term.

$$\begin{array}{rclll} && \\ \displaystyle \displaystyle \frac... ...\pm \sqrt{\displaystyle \frac{5}{24}} \\ && \\ && \end{array}$$

If we restrict the domain of the cosine function to $0\leq 3x\leq \pi \rightarrow 0\leq x\leq \displaystyle \displaystyle \frac{\pi }{3}$, we can use the arccosine function to solve for the reference angle $x^{\prime }$, and then x. The reference angle is always in the first quadrant.

$$\begin{array}{rclll} && \\ \cos \left( 3x\right) &=&\sqrt{\d... ...playstyle \frac{5}{24}}\right) \\ &=&0.3656038 \\ \end{array}$$

The period of $\cos \left( x\right)$ is $2\pi$, and the period of $\cos \left( 3x\right)$ is $$\displaystyle \frac{2\pi }{3}.$$ Divide the period into four intervals: $\left[ 0,\displaystyle \displaystyle \frac{\pi }{6}\right] ,$ $\left[ \displaystyle \displaystyle \frac{\pi }{6},\displaystyle \displaystyle \frac{% \pi }{3}\right] ,$ $\left[ \displaystyle \displaystyle \frac{\pi }{3},\displaystyle \displaystyle \frac{\pi }{2}\right] ,$ and $% \left[ \displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{2\pi }{3}\right] .$ Think of these intervals like you would think of ed quadrants I, II, III, and IV.

One solution is the angle, $x_{1}=x^{\prime }=\displaystyle \displaystyle \frac{1}{3} \cos ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{5}{24}}\right) \approx 0.3656038$ that terminates in the first quadrant. A second solution is the angle $x_{2}=\displaystyle \displaystyle \frac{\pi }{3}-x^{\prime }=% \displaystyle \d... ...left( \sqrt{\displaystyle \displaystyle \frac{5}{24}}\right) \approx 0.68159375$ that terminates in the second quadrant. A third solution is the angle $x_{3}=\displaystyle \displaystyle \frac{\pi }{3}+x^{\prime }=\displaystyle \dis... ...left( \sqrt{\displaystyle \displaystyle \frac{5}{24}}\right) \approx 1.41280135$ that terminates in the third quadrant. A fourth solution is the angle $x_{4}=% \displaystyle \displaystyle \frac{2\pi }{3}-x^{\prime }=\displaystyle \... ...left( \sqrt{\displaystyle \displaystyle \frac{5}{24}}\right) \approx 1.7287913$ that terminates in the fourth quadrant.

Since the period is $$\displaystyle \frac{2\pi }{3},$$ this means that the rest of the solutions can be found by adding or subtracting multiples of $$\displaystyle \frac{2\pi }{3% }$$ to each of the above solutions. Therefore, the solutions are

$$\begin{array}{rclll} && \\ x &=&\displaystyle \displaystyle ... ... \displaystyle \frac{2\pi }{3}\right) \\ && \\ && \end{array}$$

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=0.3656038

Left Side:

$$\displaystyle \frac{3}{5}\cos ^{2}\left( 3x\right) -\displays... ...ht) -\displaystyle \displaystyle \frac{% 1}{8}\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 0.3656038for x, then 0.3656038 is a solution.

Check the answer . x=0.68159375

Left Side:

$$\displaystyle \frac{3}{5}\cos ^{2}\left( 3x\right) -\displays... ...ght) - \displaystyle \displaystyle \frac{1}{8}\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 0.68159375for x, then 0.68159375 is a solution.

Check the answer . x=1.41280135

Left Side:

$$\displaystyle \frac{3}{5}\cos ^{2}\left( 3x\right) -\displays... ...ght) - \displaystyle \displaystyle \frac{1}{8}\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 1.41280135for x, then 1.41280135 is a solution.

Check the answer . x=1.7287913

Left Side:

$$\displaystyle \frac{3}{5}\cos ^{2}\left( 3x\right) -\displays... ...ht) -\displaystyle \displaystyle \frac{% 1}{8}\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 1.7287913for x, then 1.7287913 is a solution.

Graphical Check:

Graph the equation $f(x)=\displaystyle \displaystyle \frac{3}{5}\cos ^{2}\left( 3x\right) -\displaystyle \displaystyle \frac{1}{8}.$ (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.

Check to see if the graph crosses at the four solutions. It does. Check also to see if the graph crosses to the right and left of each solution by $% \displaystyle \displaystyle \frac{2\pi }{3}\approx 2.094395$. It does.

If you would like to review problem 9.4d, click on problem 9.4d.

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