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 $$\frac{11}{5}x$$ 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 $$\frac{11}{5}x$$ 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 $$\frac{1}{5}x.$$

$$\begin{eqnarray*}&& \\ \sin \left( \pi -\displaystyle \frac{11}{5}x\right) &=&\... ...{5}x &=&\sin ^{-1}\left( \displaystyle \frac{2}{7}\right) \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&&\\ \displaystyle \frac{11}{5}x &=&\pi -\sin ^{-1}\left( \dis... ...&& \\ x &\approx &1.29629134189\ \mbox{ radians } \\ && \\ && \end{eqnarray*}$$

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 $$\frac{10}{ 11}\pi .$$ This means that the values will repeat every $$\frac{10}{11}\pi$$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

• Left Side: $\qquad \sin \left( \displaystyle \frac{11}{5}x\right) -\displaystyle \frac{2}{7... ...ft( 0.131705318835\right) \right) -\displaystyle \frac{2}{7} \approx 0\bigskip$

• Right Side:        $0\bigskip$

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

• Left Side: $\qquad \sin \left( \displaystyle \frac{11}{5}x\right) -\displaystyle \frac{2}{7... ...left( 1.29629134189\right) \right) -\displaystyle \frac{2}{7}\approx 0\bigskip$

• Right Side:        $0\bigskip$

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 $$\frac{10}{11} \pi \approx 2.85599$$, 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 $$\frac{10}{11} \pi \approx 2.85599$$, 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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