SOLVING TRIGONOMETRIC EQUATIONS

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

Solve for x in the following equation.

Example 3:         $5\tan ^{2}\left( x\right) +6\sec \left( x\right) -1=0$

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 $\tan ^{2}\left( x\right)$ term with $\sec ^{2}\left( x\right) -1$, all the trigonometric terms will be secant terms.

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

$$\begin{array}{rclll} && \\ 5\tan ^{2}\left( x\right) +6\sec ... ...eft( x\right) +6\sec \left( x\right) -6 &=&0 \\ && \end{array}$$

Isolate the secant term. Since the left side of the equation is not easily factored, we can solve for $\sec x$ using the Quadratic Formula.

$$\begin{array}{rclll} && \\ 5\sec ^{2}\left( x\right) +6\sec ... ...-3+\sqrt{39}}\approx 1.540833>1\rightarrow \phi \\ \end{array}$$

and

$$\begin{array}{rclll} \cos x &=&\displaystyle \frac{5}{-3-\sqrt{39}}\approx -0.540833 \\ && \\ && \end{array}$$

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 graph of 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 the equation.

$$\begin{array}{rclll} (1)\qquad \cos x &=&\displaystyle \frac{... ...-3-\sqrt{39}}\right) \approx 2.142223 \\ && \\ && \end{array}$$

The angle x is the reference angle. We know that

$$\begin{array}{rclll} && \\ \cos x &=&\cos \left( -x\right) \\ && \end{array}$$

Therefore, if $\cos x=\displaystyle \displaystyle \frac{7}{-1-\sqrt{39}}$, then $\cos \left( -x\right) =% \displaystyle \displaystyle \frac{7}{-1-\sqrt{39}}.$

$$\begin{array}{rclll} && \\ \cos \left( -x\right) &=&\display... ...3-\sqrt{39}}\right) \approx -2.142223 \\ && \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ x &=&\pm \cos ^{-1}\left( \displaystyle \frac{5}{-3-\sqrt{39}}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x &\approx &\pm 2.142223\pm 6.2831853n \\ && \end{array}$$

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=2.142223

Left Side:

$$\begin{array}{rclll} 5\tan ^{2}\left( x\right) +6\sec \left( ... ...e \frac{6}{\cos \left( 2.142223\right) }-1\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 2.142223 for x, then 2.142223 is a solution.

Check solution x=-2.142223

Left Side:

$$\begin{array}{rclll} 5\tan ^{2}\left( x\right) +6\sec \left( ... ... \frac{6}{\cos \left( -2.142223\right) }-1\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -2.142223 for x, then -2.142223 is a solution.

We have just verified algebraically that the exact solutions are $x=\pm \cos ^{-1}\left( \displaystyle \displaystyle \frac{5}{-3-\sqrt{39}}\right) .$ and these solutions repeat every $\pm 2\pi$ units. The approximate values of these solutions are $% x\approx \pm 2.142223$ and these solutions repeat every $\pm 6.2831853$ units .

Graphical Check:

Graph the equation $f(x)=5\tan ^{2}\left( x\right) +6\sec \left( x\right) -1.$ 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 -2.142223. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at -2.142223+6.2831853=4.1409623 and at $-2.142223+2\left( 6.2831853\right) =10.4241476$, etc.

Verify the graph crosses the x-axis at 2.142223. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 2.142223+6.2831853=8.425408 and at $2.142223+2\left( 6.2831853\right) =14.708594$, 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 $\left[ 0,2\pi \right] :$ $x\approx 2.142223$ and $4.1409623.\bigskip\bigskip\bigskip\bigskip$

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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