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.



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