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 cross 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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Author: Nancy Marcus

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