SOLVING TRIGONOMETRIC EQUATIONS

Note:

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Solve for the real number x in the following equation.

Problem 9.3a:         $8\sin\left( x\right) -3=15\sin \left( x\right)$

Answers:        There are an infinite number of solutions: $x=\pi +\sin ^{-1}\left( \displaystyle \displaystyle \frac{3}{7}\right) \pm n\left( 2\pi \right)$ and $x=2\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{3}{7}\right) \pm n\left( 2\pi \right)$ are the exact solutions, and $x\approx 3.5845037\pm n\left( 2\pi \right)$ and $x\approx 5.840274\pm n\left( 2\pi \right)$ are the approximate solutions.

Solution:

To solve for x, first isolate the sine term.

$$\begin{eqnarray*}&& \\ 8\sin\left( x\right) -3 &=&15\sin \left( x\right) \\ &&... ...&& \\ \sin \left( x\right) &=&-\displaystyle \frac{3}{7} \\ && \end{eqnarray*}$$

If we restrict the domain of the cosine function to $-\displaystyle \displaystyle \frac{\pi }{2}\leq x\leq \displaystyle \displaystyle \frac{\pi }{2}$, we can use the arcsin function to solve for x.

$$\begin{eqnarray*}\sin\left( x\right) &=&-\displaystyle \frac{3}{7} \\ && \\ \s... ...mbox{ Reference Angle } &:&x^{\prime }=0.442911044 \\ && \\ && \end{eqnarray*}$$

The sine of x is negative in the third quadrant and the fourth quadrant. This means that there are two solutions in the first counterclockwise rotation from 0 to $2\pi$.

One angle, $x_{1}=\pi +x^{\prime }=\pi +\sin ^{-1}\left( \displaystyle \frac{3}{7}\right) \approx 3.58450369766$ terminates in the third quadrant and angle $x_{2}=2\pi -\sin ^{-1}\left( \displaystyle \frac{3}{7}\right) \approx 5.840274$ terminates in the fourth quadrant.

Since the period is $2\pi ,$ this means that the values will repeat every $2\pi$ radians. Therefore, the solutions are $x\approx 3.58450369766\pm n\left( 2\pi \right)$ and $x\approx 5.840274\pm n\left( 2\pi \right)$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 the answer . x=3.58450369766

Left Side: $\qquad 8\sin\left( x\right) -3=8\sin \left( 3.584503697667\right) -3\approx -6.42857\bigskip$

Right Side:         $15\sin \left( x\right) \approx 15\sin \left( 3.58450369766\right) \approx -6422857\bigskip$

Since the left side equals the right side when you substitute 3.58450369766for x, then 3.58450369766 is a solution.

Check the answer . x=5.840274

Left Side: $\qquad 8\sin\left( x\right) -3=8\sin \left( 5.840274\right) -3\approx -6.42857\bigskip$

Right Side:         $15\sin \left( x\right) \approx 15\sin \left( 5.840274\right) \approx -6422857\bigskip$

Since the left side equals the right side when you substitute 5.840274 for x, then 5.840274 is a solution.

Graphical Check: Graph the equation $f(x)=-7\sin \left( x\right) -3.$ (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 3.584503697667 ( one of the solutions ). Since the period of the function is $2\pi \approx 6.2831853$, the graph crosses again at 3.584503697667+6.2831853=9.867689 and again at $3.584503697667+2\left( 6.2831853\right) \approx 16.15087$, etc.

The graph also crosses at 5.840274 ( another solution we found ). Since the period is $2\pi \approx 6.2831853$, it will crosses again at 5.840274+6.2831853=12.123459 and at $5.840274+2\left( 6.2831853\right) =18.406644$, etc $.\bigskip\bigskip$

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

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