SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.11c: Solve for x in the equation

$\begin{displaymath}\displaystyle \frac{1-\sin x}{\cos x}+\displaystyle \frac{2\cos x}{1+\sin x}=8\cos x\end{displaymath}$

Answer: The exact solutions are

$\begin{displaymath}\begin{array}{rclll} x_{1} &=&\sin ^{-1}\left( -\displaystyle... ... -\displaystyle \frac{5}{8}\right) \pm 2n\pi \\ && \end{array}\end{displaymath}$

where n is an integer.

The approximate values of these solutions are

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &\approx &-0.67513153\pm 6.... ... \\ x_{2} &\approx &3.816724\pm 6.28311853n \\ && \end{array}\end{displaymath}$

where n is an integer.

Solution:

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.

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

$\begin{displaymath}\begin{array}{rclll} && \\ \displaystyle \displaystyle \frac... ...+5\right) \left( \sin x-1\right) &=&0 \\ && \\ && \end{array}\end{displaymath}$

The product of factors equals zero if at least one of the factors equals zero. If $,\ \left( 8\sin x+5\right) \left( \sin x-1\right) =0,$ then either $8\sin x+5=0$ or $\sin x-1=0\rightarrow \sin x=-\displaystyle \displaystyle \frac{5}{8}.$

$\begin{displaymath}\begin{array}{rclll} 8\sin x+5 &=&0 \\ && \\ \sin x &=&-\di... ... && \\ \sin x-1 &=&0 \\ && \\ \sin x &=&1 \\ && \end{array}\end{displaymath}$

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

$\begin{displaymath}\begin{array}{rclll} 1)\qquad \sin x &=&-\displaystyle \displ... ...^{-1}\left( 1\right) \approx 1.570796 \\ && \\ && \end{array}\end{displaymath}$

The value 1.570796 is in the list of excluded solutions. We know that $% \sin x=\sin \left( \pi -x\right) .\$ Therefore, if $\sin x=-\displaystyle \displaystyle \frac{5}{8},\$ then $\sin \left( \pi -x\right) =-\displaystyle \displaystyle \frac{5}{8}.$

$\begin{displaymath}\begin{array}{rclll} && \\ \sin \left( \pi -x\right) &=&-\di... ...e \frac{5}{8}\right) \approx 3.816724 \\ && \\ && \end{array}\end{displaymath}$

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

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &=&\sin ^{-1}\left( -\displ... ...e \displaystyle \frac{5}{8}\right) \pm 2n\pi \\ && \end{array}\end{displaymath}$

where n is an integer.

The approximate values of these solutions are

$\begin{displaymath}\begin{array}{rclll} && \\ x_{1} &\approx &-0.67513153\pm 6.... ...& \\ x_{1} &\approx &3.816724\pm 6.2831853n \\ && \end{array}\end{displaymath}$

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 $x=\sin ^{-1}\left( -\displaystyle \displaystyle \frac{5}{8}\right) \approx -0.67513153$

Left Side:

$\begin{displaymath}\begin{array}{rclll} \displaystyle \displaystyle \frac{1-\sin... ...left( -0.67513153\right) }\approx 6.24499797 \\ && \end{array}\end{displaymath}$

Right Side: $8\cos \left( -0.67513153\right) \approx 6.24499801\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -0.67513153 for x, then -0.67513153is a solution.

Check solution $x=\pi -\sin ^{-1}\left( -\displaystyle \displaystyle \frac{5}{8}\right) \approx 3.816724$

Left Side:

$\begin{displaymath}\begin{array}{rclll} \displaystyle \frac{1-\sin x}{\cos x}+\d... ...\left( 3.816724\right) } \approx -6.24499797 \\ && \end{array}\end{displaymath}$

Right Side: $8\cos \left( 3.816724\right) \approx -6.24499801\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 3.816724 for x, then 3.816724 is a solution.

Graphical Check:

Graph the equation $f(x)=\displaystyle \displaystyle \frac{1-\sin x}{\cos x}+\displaystyle \displaystyle \frac{2\cos x}{1+\sin x}% -8\cos x.$ This function is 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 -0.67513153. Since the period is $% 2\pi \approx 6.2831853$ , you can verify that the graph also crosses the x-axis again at -0.67513153+6.2831853=5.608054 and at $-0.67513153+2\left( 6.2831853\right) =11.891239$ , etc.

Verify the graph crosses the x-axis at 3.816724. Since the period is $2\pi \approx 6.2831853$ , you can verify that the graph also crosses the x-axis again at $\ 3.816724+6.2831853=9.05957$ and at $3.816724+2\left( 6.2831853\right) =16.38309$ , 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 $\left[ 0,2\pi \right] :$ $x\approx 3.816724$ and $5.608054.\bigskip\bigskip\bigskip\bigskip$

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Author: Nancy Marcus