SOLVING TRIGONOMETRIC EQUATIONS


Note: If you would like a review of trigonometry, click on trigonometry.


Problem 9.4a:        Solve for x in the following equation.



\begin{displaymath}3\sin ^{2}x-2=0\end{displaymath}

Answers:        There are an infinite number of solutions: $x=\sin ^{-1}\sqrt{%
\displaystyle \displaystyle \frac{2}{3}}\pm n\left( 2\pi \right) ,$ $x=\pi -\sin ^{-1}\sqrt{\displaystyle \displaystyle \frac{2}{3}%
}\pm n\left( 2\pi \right) ,$ $x=\pi +\sin ^{-1}\sqrt{\displaystyle \displaystyle \frac{2}{3}}\pm
n\left( 2\pi \right) ,$ and $x=2\pi -\sin ^{-1}\sqrt{\displaystyle \displaystyle \frac{2}{3}}\pm
n\left( 2\pi \right) $ are the exact solutions, and $x\approx 0.9553166\pm
n\left( 2\pi \right) ,$ $x\approx 2.186276\pm n\left( 2\pi \right) ,$ $%
x\approx 4.096909\pm n\left( 2\pi \right) ,$ and $x\approx 5.327869\pm
n\left( 2\pi \right) $ are the approximate solutions.



Solution:


To solve for x, first isolate the sine term.


\begin{displaymath}\begin{array}{rclll}
&& \\
3\sin ^{2}x-2 &=&0 \\
&& \\
\si...
...ght) &=&\pm \sqrt{\displaystyle \frac{2}{3}} \\
&&
\end{array}\end{displaymath}

If we restrict the domain of the sine function to $-\displaystyle \displaystyle \frac{\pi }{2}\leq
x\leq \displaystyle \displaystyle \frac{\pi }{2}$, we can use the arcsin function to solve for the reference angle $x^{\prime }$, and then x. The reference angle is always in the first quadrant.

\begin{displaymath}\begin{array}{rclll}
&& \\
\sin \left( x\right) &=&\sqrt{\di...
...{2}{3}}\right) \\
&& \\
x &\approx &0.9553166 \\
\end{array}\end{displaymath}

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


One angle, $x_{1}=x^{\prime }=\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right)
\approx 0.9553166$ terminates in the first quadrant, a second angle $%
x_{2}=\pi -x^{\prime }=\pi -\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right)
\approx 2.186276$ terminates in the second quadrant, a third angle $%
x_{3}=\pi +x^{\prime }=\pi +\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right)
\approx 4.096909$ terminates in the third quadrant, and a fourth angle $%
x_{4}=2\pi -x^{\prime }=2\pi -\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right)
\approx 5.327869$ 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

\begin{displaymath}\begin{array}{rclll}
&& \\
x &=&\sin ^{-1}\left( \sqrt{\disp...
...}{3}}\right) \pm n\left( 2\pi
\right) \\
&& \\
&&
\end{array}\end{displaymath}

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=0.9553166


Left Side:

\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 0.9553166\right) -2\approx
0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side equals the right side when you substitute 0.9553166for x, then 0.9553166 is a solution.




Check the answer . x=2.186276


Left Side:

\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 2.186276\right) -2\approx
0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side equals the right side when you substitute 2.186276 for x, then 2.186276 is a solution.




Check the answer . x=4.096909


Left Side:

\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 4.096909\right) -2\approx
0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side equals the right side when you substitute 4.096909 for x, then 4.096909 is a solution.




Check the answer . x=5.327869


Left Side:

\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 5.327869\right) -2\approx
0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side equals the right side when you substitute 5.327869 for x, then 5.327869 is a solution.




Graphical Check:


Graph the equation $f(x)=3\sin ^{2}x-2.$ (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 0.9553166, 2.186276, 4.096909, and 5.327869 . Since the period is $2\pi \approx 6.283185$, the graph will cross the x-axis every $2\pi $ units to the left and right of each number.


If you would like to review problem 9.4b, click on problem 9.4b.

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Author: Nancy Marcus

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