SOLVING TRIGONOMETRIC EQUATIONS

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

Example 1:        Solve for x in the following equation.

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

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. $\cos x\neq 0\rightarrow x\neq \pm \displaystyle \frac{\pi }{2}\pm n\pi ,\qquad$and         $\sin x\neq -1\rightarrow x\neq \displaystyle \frac{3\pi }{2}\pm 2n\pi .$ Therefore, before we even start to solve the problem, the set of real numbers in the set $\left\{ \pm \displaystyle \frac{\pi }{2}\pm n\pi \right\}$ must be excluded from the possible set of solutions.

To simplify the equation, let's multiple the second fraction by 1 in the form $$\frac{1-\sin x}{1-\sin x}$$, then 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. Be sure to check your answers with the original equation.

$$\begin{array}{rclll} && \\ \displaystyle \frac{1+\sin x}{\co... ...&=&4 \\ && \\ \cos x &=&\frac{1}{2} \\ && \\ && \end{array}$$

How do we isolate the x in this equation? 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.

$$\begin{array}{rclll} \cos x &=&\frac{1}{2} \\ && \\ \cos ^... ...\frac{1}{2}\right) \approx 1.04719755 \\ && \\ && \end{array}$$

We know that $\cos x=\cos \left( -x\right) .\$Therefore, if $\cos x=\displaystyle \frac{1% }{2},\$then $\cos \left( -x\right) =\displaystyle \frac{1}{2}.$

$$\begin{array}{rclll} && \\ \cos \left( -x\right) &=&\display... ...frac{1}{2}\right) \approx -1.04719755 \\ && \\ && \end{array}$$

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

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

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x &\approx &\pm 1.04719755\pm 6.2831853n \\ && \end{array}$$

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=1.04719755

Left Side:

$$\frac{1+\sin x}{\cos x}+ \displaystyle \frac{\cos x}{1+\sin x... ...1.04719755\right) }{1+\sin \left( 1.04719755\right) }\approx 4$$

Right Side:        $4\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 1.04719755 for x, then 1.04719755is a solution.

Check solution x=-1.04719755

Left Side:

$$\begin{array}{rclll} \displaystyle \frac{1+\sin x}{\cos x}+\d... ...{1+\sin \left( -1.04719755\right) }\approx 4 \\ && \end{array}$$

Right Side:        $4\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -1.04719755 for x, then -1.04719755is a solution.

We have just verified algebraically that the exact solutions are $x=\pm \cos ^{-1}\left( \displaystyle \frac{1}{2}\right)$ and these solutions repeat every $\pm 2\pi$ units. The approximate values of these solutions are $x\approx \pm 1.04719755$ and these solutions repeat every $\pm 6.2831853$ units.

Graphical Check:

Graph the function $f(x)=\displaystyle \frac{1+\sin x}{\cos x}+\displaystyle \frac{\cos x}{1+\sin x}-4$ , 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. Let's check a few of these x-intercepts against the solutions we derived.

Verify the graph crosses the x-axis at 1.04719755. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 1.04719755+6.2831853=7.330383 and at $1.04719755+2\left( 6.2831853\right) =13.613568$, etc.

Verify the graph crosses the x-axis at -1.04719755. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at -1.04719755+6.2831853=5.23599 and at $1.04719755+2\left( 6.2831853\right) =$11.519173, 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 $\left[ 0,2\pi \right] :$ $x\approx 1.04719755$ and $5.23599.\bigskip\bigskip\bigskip\bigskip$

If you would like to work another example, click on Example.

If you would like to test yourself by working some problems similar to this example, click on Problem.

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