SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.12c: Solve for x in the equation

$\begin{displaymath}6\cot ^{4}x-8\cot ^{3}x-31\cot^{2}x+17\cot x-2=0\end{displaymath}$

Answer: The exact answers are

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &=&\tan ^{-1}\left( \displa... ...playstyle \frac{1}{2}\right) \pm n\pi \\ && \\ && \end{array}\end{displaymath}$

where n is an integer.

The approximate values of these solutions are

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &\approx &0.3404548\pm 3.14... ... x_{4} &\approx &-0.46364776\pm 3.141592653n \\ && \end{array}\end{displaymath}$

$\quad$ where n is an integer.

Original Equation: $\qquad 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\cot x-2=0\bigskip$

Solution:

There are an infinite number of solutions to this problem. Let's simplify the problem by rewriting it in an equivalent factored form.

$\begin{displaymath}\begin{array}{rclll} && \\ 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{... ...ht) \left( 3\cot ^{2}x+5\cot x-2\right) &=&0 \\ && \end{array}\end{displaymath}$

The only way the product equals zero is if at least one of the factors equals zero. Therefore,

$\begin{displaymath}6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\cot x-24 =0\end{displaymath}$

$\begin{displaymath}\begin{array}{rclll} \left( 1\right) \qquad 2\cot ^{2}x-6\cot... ... 2\right) \qquad 3\cot ^{2}x+5\cot x-2 &=&0. \\ && \end{array}\end{displaymath}$

Let's first solve for $\cot x$ in equations $\left( 1\right)$ and $\left( 2\right) .$

$\begin{displaymath}\begin{array}{rclll} && \\ \left( 1\right) \qquad 2\cot ^{2}... ...{5\pm 7}=-\displaystyle \frac{1}{2},3 \\ && \\ && \end{array}\end{displaymath}$

How do we isolate the x in each of these equations? We could take the inverse (arctangent) of both sides of the equation. However, the tangent function is not a one-to-one function.

Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. The tangent function is one-to-one on the interval $\left( -\displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{\pi }{2}\right) .$ If we restrict the domain of the tangent function to that interval , we can take the arctangent of both sides of each equation.

$\begin{displaymath}\begin{array}{rclll} \left( 1\right) \qquad \tan x &=&\displa... ...\frac{1}{2}\right) \approx -0.4636476 \\ && \\ && \end{array}\end{displaymath}$

Since the period of $\tan x$ equals $\pi$ , these solutions will repeat every $\pi$ units. The exact solutions are

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &=&x_{1}=\tan ^{-1}\left( \... ...( -\displaystyle \frac{1}{2}\right) \pm n\pi \\ && \end{array}\end{displaymath}$

where n is an integer.

The approximate values of these solutions are

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &\approx &0.3404548\pm 3.14... ... x_{4} &\approx &-0.4636476\pm 3.141592653n \\ && \end{array}\end{displaymath}$

where n is an integer.

Original Equation: $\qquad 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\cot x-2=0\bigskip$

You can check each solution algebraically by substituting each solution in the original equation. If, after the substitution, the left side of the original equation equals the right side of the original equation, the solution is valid.

You can also check the solutions graphically by graphing the function formed by subtracting the right side of the original equation from the left side of the original equation. The solutions of the original equation are the x-intercepts of this graph.

Algebraic Check:

Check solution $\qquad x_{1}=\tan ^{-1}\left( \displaystyle \displaystyle \frac{2}{3+\sqrt{7}}\right) \approx 0.3404548$

Left Side:

$\begin{displaymath}\begin{array}{rclll} 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\... ...laystyle \frac{17}{\tan \left( 0.3404548\right) }-2 \end{array}\end{displaymath}$

Right Side: $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.72972766 for x, then 0.72972766is a solution.

Check solution $\qquad x_{2}=\tan ^{-1}\left( \displaystyle \displaystyle \frac{2}{3-\sqrt{7}}\right) \approx 1.395490$

Left Side:

$\begin{displaymath}\begin{array}{rclll} 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\... ...playstyle \frac{17}{\tan \left( 1.395490\right) }-2 \end{array}\end{displaymath}$

Right Side: $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 1.395490 for x, then 1.395490 is a solution.

Check solution $\qquad x_{3}=\tan ^{-1}\left( 3\right) \approx 1.2490577$

Left Side:

$\begin{displaymath}\begin{array}{rclll} 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\... ...laystyle \frac{17}{\tan \left( 1.2490577\right) }-2 \end{array}\end{displaymath}$

Right Side: $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 1.2490577 for x, then 1.2490577 is a solution.

Check solution $\qquad x_{4}=\tan ^{-1}\left( -\displaystyle \frac{1}{2}\right) \approx -0.4636476$

Left Side:

$\begin{displaymath}\begin{array}{rclll} 6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\... ...aystyle \frac{17}{\tan \left( -0.4636476\right) }-2 \end{array}\end{displaymath}$

Right Side: $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -0.4636476 for x, then -0.4636476is a solution.

Graphical Check:

Graph the function $f(x)=6\cot ^{4}x-8\cot ^{3}x-31\cot ^{2}x+17\cot x-2$ , formed by subtracting the right side of the original equation from the left side of the original equation. The x-intercepts are the real solutions.

. Note that the graph crosses the x-axis many times indicating many solutions. Let's check a few of these x-intercepts against the solutions we derived.

Verify the graph crosses the x-axis at 0.3404548. Since the period is $\pi \approx 3.141592653$ , you can verify that the graph also crosses the x-axis again at 0.3404548+3.141592653=3.482047 and at $0.3404548+2\left( 3.141592653\right) =6.62364$ , etc.

Verify the graph crosses the x-axis at 1.39549. Since the period is $\pi \approx 3.141592653$ , you can verify that the graph also crosses the x-axis again at 1.39549+3.141592653=4..5370827 and at $1.39549+2\left( 3.141592653\right) =7.678675$ , etc.

Verify the graph crosses the x-axis at 1.2490577. Since the period is $\pi \approx 3.141592653$ , you can verify that the graph also crosses the x-axis again at 1.2490577+3.141592653=4.39065 and at $1.2490577+2\left( 3.141592653\right) =7.53224,$ etc.

Verify the graph crosses the x-axis at -0.4636476. Since the period is $% \pi \approx 3.141592653$ , you can verify that the graph also crosses the x-axis again at -0.4636476+3.141592653=2.677945 and at $-0.4636476+2\left( 3.141592653\right) =5.8195377$ , etc.

Note: If the problem were to find the solutions in the interval $\left[ 0,2\pi \right]$ , then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,2\pi \right] :$ $x\approx 0.3404548,\ 1.2490577,\ 1.39549,\ 2.677945,\ 3.482047,$ 4.39065, 4.5370827 and $5.819538.\bigskip\bigskip\bigskip\bigskip$

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Author: Nancy Marcus