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.

If you would like to go to the next section, click on Next.


If you would like to go back to the previous section, click on previous.

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


[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]

S.O.S MATHematics home page



Author: Nancy Marcus

Copyright © 1999-2024 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour