SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.11b:        Solve for x in the equation

$$\displaystyle \frac{1-4\cos x}{2\sin x}+\displaystyle \frac{2\sin x}{1-\cos x}=7$$

Answer:    The exact answers are

$$\begin{array}{rclll} && \\ x_{1} &=&\sin ^{-1}\left( \displa... ... \displaystyle \frac{5}{14}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.36520722\pm 6.2... ... \\ x_{2} &\approx &2.7763854\pm 6.2831853n \\ && \end{array}$$

$\quad$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. $\sin x\neq 0\rightarrow x\neq \pm n\pi ,\qquad$and         $\cos x\neq 1\rightarrow x\neq \pm 2n\pi .$ Therefore, before we even start to solve the problem, the set of real numbers in the set $\left\{ \pm n\pi \right\}$ must be excluded from the possible set of solutions.

The equation has two different trigonometric terms. Let's manipulate the equation so that we have an equation of like trigonometric terms.

First, let's multiple the second fraction by 1 in the form $$\displaystyle \frac{1+\cos x}{% 1+\cos x}.$$

$$\begin{array}{rclll} && \\ \displaystyle \displaystyle \frac... ... \sin x &=&\displaystyle \frac{5}{14} \\ && \\ && \end{array}$$

How do we isolate the x in this equation? We could take the arcsine of both sides of the equation. However, the sine 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 sine 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 sine function to that interval , we can take the arcsine of both sides of the equation.

$$\begin{array}{rclll} \sin x &=&\displaystyle \frac{5}{14} \\ ... ... \frac{5}{14}\right) \approx 0.365207 \\ && \\ && \end{array}$$

We know that $\sin x=\sin \left( \pi -x\right) .\$Therefore, if $\sin x=% \displaystyle \displaystyle \frac{5}{14},\$then $\sin \left( \pi -x\right) =\displaystyle \displaystyle \frac{5}{14}.$

$$\begin{array}{rclll} && \\ \sin \left( \pi -x\right) &=&\dis... ...\frac{5}{14}\right) \approx 2.7763854 \\ && \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ x_{1} &=&\sin ^{-1}\left( \displa... ... \displaystyle \frac{5}{14}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.365207\pm 6.283... ... \\ x_{2} &\approx &2.7763854\pm 6.2831853n \\ && \end{array}$$

where n is an integer.

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 $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{5}{14}\right) \approx 0.365207$

Left Side:

$$\begin{array}{rclll} \displaystyle \frac{1-4\cos x}{2\sin x}+... ....365207\right) }{1-\cos 0.365207}% \approx 7 \\ && \end{array}$$

Right Side:        $7\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.365207 for x, then 0.365207 is a solution.

Check solution $x=\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{5}{14}\right) \approx 2.7763854$

Left Side:

$$\begin{array}{rclll} \displaystyle \frac{1-4\cos x}{2\sin x}+... ... }{1-\cos \left( 2.7763854\right) }\approx 7 \\ && \end{array}$$

Right Side:        $7\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 2.7763854 for x, then 2.7763854 is a solution.

The exact solutions are $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{5}{14}\right)$ and $% x=\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{5}{14}\right)$ and these solutions repeat every $\pm 2\pi$ units. The approximate values of these solutions are $% x\approx 0.365207$and $x\approx 2.7763854$ and these solutions repeat every $% \pm 6.2831853$ units.

Graphical Check:

Graph the function $f(x)=\displaystyle \displaystyle \frac{1-4\cos x}{2\sin x}+\displaystyle \displaystyle \frac{2\sin x}{1-\cos x}% -7,$ formed by subtracting the right side of the original equation from the left side of the original equation. The x-intercepts are the solutions to the original equation.

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 0.365207. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 0.365207+6.2831853=7.301301 and at $0.365207+2\left( 6.2831853\right) =13.8844866$, etc.

Verify the graph crosses the x-axis at 2.7763854. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at $\ 2.7763854+6.2831853=9.05957$ and at $2.7763854+2\left( 6.2831853\right) =$15.342756, 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 $\left[ 0,2\pi \right] :$ $x\approx 0.365207$ and $2.7763854.\bigskip\bigskip\bigskip\bigskip$

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