SOLVING TRIGONOMETRIC EQUATIONS


Note:

If you would like an review of trigonometry, click on trigonometry.



Solve for x in the following equation.


Example 1:

$\sin\left( x\right) +\sqrt{2}=-\sin x$



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( x\right) +\sqrt{2} &=&-\sin x \\
&& \\
2\si...
...right) &=&-\displaystyle \frac{\sqrt{2}}{2} \\
&& \\
&& \\
&&
\end{eqnarray*}


If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2}
\leq x\leq ,\displaystyle \frac{\pi }{2}\right] $, we can use the inverse sine function to solve for reference angle x, and then x.

\begin{eqnarray*}&& \\
\sin \left( x\right) &=&-\displaystyle \frac{\sqrt{2}}{2...
...sqrt{2}}{2}\right) \\
&& \\
x &\approx &-0.785398163397 \\
&&
\end{eqnarray*}
\begin{eqnarray*}&&\\
\mbox{ Reference Angle } &:&x=\sin ^{-1}\left( \displayst...
...ngle } &:&x^{\prime }\approx 0.785398163397 \\
&& \\
&& \\
&&
\end{eqnarray*}


We know that the $\sin $e function is negative in the third and the fourth quadrant. Therefore two of the solutions are the angle $\pi +x^{\prime }$that terminates in the third quadrant and the angle $2\pi -x^{\prime }$ that terminates in the fourth. We have already solved for $x^{\prime }.$

\begin{eqnarray*}&& \\
\mbox{ Third Quadrant } &:&x_{1}=\pi +x^{\prime }=\pi +\...
...{2}}{2}\right) \\
&& \\
x_{2} &\approx &5.497787 \\
&& \\
&&
\end{eqnarray*}


The solutions are $x=\pi +\sin ^{-1}\left( \displaystyle \frac{\sqrt{2}}{2}\right) $ and $
x=2\pi -\sin ^{-1}\left( \displaystyle \frac{\sqrt{2}}{2}\right) .\bigskip\bigskip
\bigskip $

The period of the sin $\left( x\right) $ function is $2\pi .$ This means that the values will repeat every $2\pi $ radians in both directions. Therefore, the exact solutions are $x=\pi +\sin ^{-1}\left( \displaystyle \frac{\sqrt{2}}{2}\right)
\pm n\left( 2\pi \right) $ and $x=2\pi -\sin ^{-1}\left( \displaystyle \frac{\sqrt{2}}{2}
\right) \pm n\left( 2\pi \right) $ where n is an integer.


The approximate solutions are $x\approx 3.92699081699\pm n\left( 2\pi
\right) $ and $x\approx 5.497787\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 answer x=3.92699081699


Since the left side equals the right side when you substitute 3.92699081699for x, then 3.92699081699 is a solution.




Check answer x=5.497787


Since the left side equals the right side when you substitute 5.497787 for x, then 5.497787 is a solution.



Graphical Check:


Graph the equation

$f(x)=\sin\left( x\right) +\sqrt{2}+\sin x.$

Note that the graph crosses the x-axis many times indicating many solutions.


Note that it crosses at 3.92699081699. Since the period is $2\pi \approx
6.2831853$, it crosses again at 3.92699081699+6.2831853=10.21017 and at 3.92699081699+2(6.2831853)=16.49336, etc.


The graph also crosses the x-axis at 5.497787. Since the period is $2\pi \approx
6.2831853$, it crosses again at 5.497787+6.2831853=11.78097 and at 5.497787+2(6.2831853)=18.06415, etc.



If you would like to work another example, click on Example.


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



If you would like to go back to the equation table of contents, click on Contents.


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Author: Nancy Marcus

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