SOLVING TRIGONOMETRIC EQUATIONS


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


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



\begin{displaymath}4\cot ^{2}x-3=0\end{displaymath}

Answers:        There are an infinite number of solutions: $x=\tan ^{-1}%
\displaystyle \frac{2}{\sqrt{3}}\pm n\left( \pi \right) $ and $x=-\tan ^{-1}\displaystyle \displaystyle \frac{2}{%
\sqrt{3}}\pm n\left( \pi \right) $ are the exact solutions, and $x\approx
0.8570719\pm n\left( \pi \right) $ and $x\approx -0.8570719\pm n\left( \pi
\right) $ are the approximate solutions.



Solution:


To solve for x, first isolate the cotangent term.


\begin{displaymath}\begin{array}{rclll}
&& \\
4\cot ^{2}x-3 &=&0 \\
&& \\
\co...
...an x &=&\pm \displaystyle \frac{2}{\sqrt{3}} \\
&&
\end{array}\end{displaymath}

If we restrict the domain of the tangent function to $-\displaystyle \displaystyle \frac{\pi }{2}<x<%
\displaystyle \displaystyle \frac{\pi }{2}$, we can use the arctan 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}
&& \\
\tan x &=&\displaystyle \frac{2}{...
...rt{3}}\right) \\
&& \\
x &\approx &0.85707194 \\
\end{array}\end{displaymath}

In the interval from $-\displaystyle \displaystyle \frac{\pi }{2}$ to $\displaystyle \displaystyle \frac{\pi }{2}$ the tangent of <tex2htmlcommentmark> x is positive in the first quadrant and negative in the fourth quadrant.


One angle, $x_{1}=x^{\prime }=\tan ^{-1}\left( \displaystyle \displaystyle \frac{2}{\sqrt{3}}\right)
\approx 0.85707194$ terminates in the first quadrant, and a fourth angle $%
x_{4}=-x^{\prime }=-\tan ^{-1}\left( \displaystyle \displaystyle \frac{2}{\sqrt{3}}\right) \approx
-0.85707194$ terminates in the fourth quadrant..


Since the period is $\pi ,$ this means that the rest of the solutions can be found by adding $\pi $ to each of the above solutions. Therefore, the solutions are

\begin{displaymath}\begin{array}{rclll}
&& \\
x &=&\tan ^{-1}\left( \displaysty...
...rt{3}}\right) \pm n\left( \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=0.85707194


Left Side:

\begin{displaymath}4\cot ^{2}x-3\approx 4\cot ^{2}\left( 0.85707194\right)
-3\approx 0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side equals the right side when you substitute 0.85707194for x, then 0.85707194 is a solution.




Check the answer . x=-0.85707194


Left Side:

\begin{displaymath}4\cot ^{2}x-3\approx 4\cot ^{2}\left( -0.85707194\right)
-3\approx 0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side equals the right side when you substitute -0.85707194for x, then -0.85707194 is a solution.




Graphical Check:


Graph the equation $f(x)=4\cot ^{2}x-3.$ (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 -0.85707194 and -0.85707194. Since the period is $\pi \approx 3.1415926$, the graph will cross the x-axis again every $\pi $ units to the left and right of each number.


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

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Author: Nancy Marcus

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