SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.11a:        Solve for x in the equation         

$$\displaystyle \frac{1-\sin x}{\cos x}+\displaystyle \frac{\cos x}{1-\sin x}=8$$

Answer:    The exact answers are

$$\begin{array}{rclll} && \\ x &=&\pm \cos ^{-1}\left( \displaystyle \frac{1}{4}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x &\approx &\pm 1.318116\pm 6.2831853n \\ && \end{array}$$

$\quad$where n is an integer.

Solution:

There are an infinite number of solutions to this problem. Since denominators of fractions cannot equal zero, real numbers that cause the denominators to equal zero must be eliminated from the set of possible solutions. $\cos x\neq 0\rightarrow x\neq \pm \displaystyle \displaystyle \frac{\pi }{2}\pm n\pi ,\qquad$and         $\sin x\neq 1\rightarrow x\neq \displaystyle \displaystyle \frac{\pi }{2}\pm 2n\pi .$Therefore, before we even start to solve the problem, the set of real numbers in the set $\left\{ \pm \displaystyle \displaystyle \frac{\pi }{2}\pm n\pi \right\}$ must be excluded from the possible set of solutions.

To simplify the equation, let's multiple the second equation by 1 in the form $$\displaystyle \frac{1+\sin x}{1+\sin x}$$, simplify and solve. The result will be an equation that may not be equivalent to the original equation, but an equation where we can solve for x. With this type of manipulation, there may be extraneous solutions. In other words, you may come up with solutions for the new equation that are not solutions to the original equation. Therefore, check your solutions with the original equation.

$$\begin{array}{rclll} && \\ \displaystyle \displaystyle \frac... ... \cos x &=&\displaystyle \frac{1}{4} \\ && \\ && \end{array}$$

How do we isolate the x in each of these equations? We could take the arccosine of both sides of the equation. However, the cosine 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 cosine function is one-to-one on the interval $\left[ 0,\pi \right] .$ If we restrict the domain of the cosine function to that interval , we can take the arccosine of both sides of each equation.

$$\begin{array}{rclll} \cos x &=&\displaystyle \frac{1}{4} \\ ... ...e \frac{1}{4}\right) \approx 1.318116 \\ && \\ && \end{array}$$

We know that $\cos x=\cos \left( -x\right) .\$Therefore, if $\cos x=\displaystyle \displaystyle \frac{1% }{4},\$then $\cos \left( -x\right) =\displaystyle \displaystyle \frac{1}{4}.$

$$\begin{array}{rclll} && \\ \cos \left( -x\right) &=&\display... ... \frac{1}{4}\right) \approx -1.318116 \\ && \\ && \end{array}$$

Since the period of $\cos x$ equals $2\pi$, these solutions will repeat every $2\pi$ units. The exact solutions are

$$\begin{array}{rclll} && \\ x &=&\pm \cos ^{-1}\left( \displaystyle \frac{1}{4}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x &\approx &\pm 1.318116\pm 6.2831853n \\ && \end{array}$$

where n is an integer.

$$\frac {} {}$$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 x=1.318116

Left Side:

$$\begin{array}{rclll} \displaystyle \frac{1-\sin x}{\cos x}+\d... ... }{1-\sin \left( 1.318116\right) } \approx 8 \\ && \end{array}$$

Right Side:        $8\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 1.318116 for x, then 1.318116 is a solution.

Check solution x=-1.318116

Left Side:

$$\begin{array}{rclll} \displaystyle \frac{1-\sin x}{\cos x}+\d... ... }{1-\sin \left( -1.318116\right) }\approx 8 \\ && \end{array}$$

Right Side:         $x=8\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -1.318116 for x, then -1.318116 is a solution.

We have just verified algebraically that the exact solutions are $x=\pm \cos ^{-1}\left( \displaystyle \displaystyle \frac{1}{4}\right)$ and these solutions repeat every $\pm 2\pi$ units. The approximate values of these solutions are $x\approx \pm 1.318116$ and these solutions repeat every $\pm 6.2831853$ units.

Graphical Check:

Graph the function $f(x)=\displaystyle \displaystyle \frac{1-\sin x}{\cos x}+\displaystyle \displaystyle \frac{\cos x}{1-\sin x}-8$ , 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 1.318116. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 1.318116+6.2831853=7.601301 and at $1.318116+2\left( 6.2831853\right) =13.8844866$, etc.

Verify the graph crosses the x-axis at -1.318116. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at -1.318116+6.2831853=4.965069 and at $-1.318116+2\left( 6.2831853\right) =$11.2482546, 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 $\left[ 0,2\pi \right] :$ $x\approx 1.318116$ and $4.965069.\bigskip\bigskip\bigskip\bigskip$

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