SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.10a:        Solve for x in the equation

$$3\sec ^{2}x+19\tan x-17=0$$

Answer:    The exact answers are

$$\begin{array}{rclll} && \\ x_{1} &=&\tan ^{-1}\left( \displa... ...x_{2} &=&\tan ^{-1}\left( -7\right) \pm n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.5880026\pm 3.14... ...\\ x_{2} &\approx &-1.428899\pm 3.14159265n \\ && \end{array}$$

$\quad$where n is an integer.

Solution:

There are an infinite number of solutions to this problem.

We can make the solution easier if we convert all the trigonometric terms to like trigonometric terms.

One common trigonometric identity is $1+\tan ^{2}(x)=\sec ^{2}\left( x\right) .$ If we replace the $\sec ^{2}\left( x\right)$ term with $1+\tan ^{2}(x)$, all the trigonometric terms will be tangent terms.

Replace $\sec ^{2}\left( x\right)$ with $1+\tan ^{2}(x)$ in the original equation and simplify.

$$\begin{array}{rclll} && \\ 3\sec ^{2}x+19\tan x-17 &=&0 \\ ... ...&& \\ 3\tan ^{2}(x)+19\tan x-14 &=&0 \\ && \\ && \end{array}$$

Isolate the tangent term. We can solve for $\tan x$ by rewriting the equation in an equivalent factored form.

$$\begin{array}{rclll} && \\ 3\tan ^{2}(x)+19\tan x-14 &=&0 \\... ...-2\right) \left( \tan x+7\right) &=&0 \\ && \\ && \end{array}$$

The product of two factors equals zero if at least one of the factors equals zeros. This means that $3\tan ^{2}(x)+19\tan x-14=0\$if $3\tan -2$ or $\tan x+7=0.$

$$\begin{array}{rclll} && \\ 3\tan x-2 &=&0 \\ && \\ \tan x &=&\displaystyle \frac{2}{3} \\ \end{array}$$

and

$$\begin{array}{rclll} \tan x+7 &=&0 \\ && \\ \tan x &=&-7 \\ && \\ && \end{array}$$

How do we isolate the x in each of these equations? We could take the inverse (arctangent) of both sides of each 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 graph of 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.

$$\begin{array}{rclll} (1)\qquad \tan x &=&\displaystyle \frac{... ...an ^{-1}\left( -7\right) \approx -1.42889927 \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ x_{1} &=&\tan ^{-1}\left( \displaystyle \frac{2}{3}\right) \pm n\pi \\ \end{array}$$

and

$$\begin{array}{rclll} x_{2} &=&\tan ^{-1}\left( -7\right) \pm n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.5880026\pm 3.14... ... x_{2} &\approx &-1.42889927\pm 3.14159265n \\ && \end{array}$$

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 $x=\tan ^{-1}\left( \displaystyle \displaystyle \frac{2}{3}\right) \approx 0.5880026$

Left Side:

$$\begin{array}{rclll} 3\sec ^{2}x+19\tan x-17 &=&\displaystyle... ...ight) }+19\tan \left( 0.5880026\right) -17\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.5880026 for x, then 0.5880026 is a solution.

Check solution $x=\tan ^{-1}\left( -7\right) \approx -1.42889927$

Left Side:

$$\begin{array}{rclll} 3\sec ^{2}x+19\tan x-17 &=&\displaystyle... ...ht) }+19\tan \left( -1.42889927\right) -17\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -1.42889927 for x, then -1.42889927is a solution.

Graphical Check:

Graph the equation $f(x)=3\sec ^{2}x+19\tan x-17\$or $\ f\left( x\right) =% \displaystyle \frac{3}{\cos ^{2}x}+19\tan x-17.$

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 0.5880026. Since the period is $% 2\pi \approx 3.14159265$, you can verify that the graph also crosses the x-axis again at 0.5880026+3.14159265=3.729595 and at $0.5880026+2\left( 3.14159265\right) =6.871188$, etc.

Verify the graph crosses the x-axis at -1.42889927. Since the period is $% 2\pi \approx 3.14159265$, you can verify that the graph also crosses the x-axis again at -1.42889927+3.14159265=1.7126934 and at $% -1.42889927+2\left( 3.14159265\right) =4.854286$, 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 $\left[ 0,2\pi \right] :$ $x\approx 0.5880026,$ 1.7126934, 3.729595, and $4.854286.\bigskip\bigskip \bigskip\bigskip$

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