SOLVING TRIGONOMETRIC EQUATIONS

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

Solve for x in the following equation.

Example 3:         $5\cot ^{2}\left( x\right) +3\csc \left( x\right) -7=0$

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 $1+\cot ^{2}(x)=\csc ^{2}\left( x\right) .$ If we replace the $\cot ^{2}\left( x\right)$ term with $\csc ^{2}\left( x\right) -1$, all the trigonometric terms will be secant terms.

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

$$\begin{array}{rclll} && \\ 5\cot ^{2}\left( x\right) +3\csc ... ...ft( x\right) +3\csc \left( x\right) -12 &=&0 \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ 5\csc ^{2}\left( x\right) +3\csc ... ...sin x &=&\displaystyle \frac{10}{-3+\sqrt{249}} \\ \end{array}$$

and

$$\begin{array}{rclll} \sin x &=&\displaystyle \frac{10}{-3-\sqrt{249}} \\ && \\ && \end{array}$$

To solve for x, we have to isolate x. How do we isolate the x? We could take the inverse (arcsine) of both sides. However, inverse functions can only be applied to one-to-one functions and the sine function is not one-to-one.

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 $\left[ -\displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{\pi }{2}% \right] .$ If we restrict the domain of the sine function to that interval , we can take the arcsine of both sides of each equation.

$$\begin{array}{rclll} (1)\qquad \sin x &=&\displaystyle \frac{... ...\sqrt{249}}\right) \approx -0.5615383 \\ && \\ && \end{array}$$

The angle x is the reference angle. We know that

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

Therefore, if $\sin x=\displaystyle \displaystyle \frac{10}{-3+\sqrt{249}}$, then sin $\left( -x\right) =\pi -\displaystyle \displaystyle \frac{10}{-3+\sqrt{249}},$ and if $\sin x=\displaystyle \displaystyle \frac{10}{-3-\sqrt{249}}$, then sin $\left( -x\right) =\pi -\displaystyle \displaystyle \frac{10}{-3-\sqrt{249}}.$

$$\begin{array}{rclll} && \\ 3)\qquad \sin \left( \pi -x\right... ...rac{10}{-3+\sqrt{249}}\right) \approx 2.2429396 \\ \end{array}$$

and

$$\begin{array}{rclll} 4)\qquad \sin \left( \pi -x\right) &=&\d... ...\sqrt{249}}\right) \approx 3.70313096 \\ && \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ x_{1} &=&\sin ^{-1}\left( \displa... ...le \frac{10}{-3-\sqrt{249}}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x_{1} &\approx &0.898653\pm 6.283... ...\\ x_{4} &\approx &3.70313096\pm 6.2831853n \\ && \end{array}$$

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 $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{10}{-3+\sqrt{249}}\right) \approx 0.898653$

Left Side:

$$\begin{array}{rclll} 5\cot ^{2}\left( x\right) +3\csc \left( ... ...e \frac{3}{\sin \left( 0.898653\right) }-7\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 0.898653 for x, then 0.898653 is a solution.

Check solution $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{10}{-3-\sqrt{249}}\right) \approx -0.5615383$

Left Side:

$$\begin{array}{rclll} 5\cot ^{2}\left( x\right) +3\csc \left( ... ...\frac{3}{\sin \left( -0.5615383\right) }-7\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -0.5615383 for x, then -0.5615383is a solution.

Check solution $x=\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{10}{-3+\sqrt{249}}\right) \approx 2.2429396$

Left Side:

$$\begin{array}{rclll} 5\cot ^{2}\left( x\right) +3\csc \left( ... ... \frac{3}{\sin \left( 2.2429396\right) }-7\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 2.2429396 for x, then 2.2429396 is a solution.

Check solution $x=\pi -\sin ^{-1}\left( \displaystyle \frac{10}{-3-\sqrt{249}}\right) \approx 3.70313096$

Left Side:

$$\begin{array}{rclll} 5\cot ^{2}\left( x\right) +3\csc \left( ... ...\frac{3}{\sin \left( 3.70313096\right) }-7\approx 0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 3.70313096 for x, then 3.70313096is a solution.

Graphical Check:

Graph the equation $f(x)=5\cot ^{2}\left( x\right) +3\csc \left( x\right) -7$or $f\left( x\right) =\displaystyle \displaystyle \frac{5}{\tan ^{2}\left( x\right) }+\displaystyle \displaystyle \frac{3}{\sin \left( x\right) }-7.$ 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 0.898653. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 0.898653+6.2831853=7.1818383 and at $0.898653+2\left( 6.2831853\right) =13.465024$, etc.

Verify the graph crosses the x-axis at -0.5615383. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at -0.5615383+6.2831853=5.721647 and at $-0.5615383+2\left( 6.2831853\right) =12.004823$, etc.

Verify the graph crosses the x-axis at 2.2429396. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 2.2429396+6.2831853=8.526125 and at $2.2429396+2\left( 6.2831853\right) =14.80931$, etc.

Verify the graph crosses the x-axis at 3.70313096. Since the period is $% 2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 3.70313096+6.2831853=9.9863163 and at $3.70313096+2\left( 6.2831853\right) =16.269502$, etc.

Note: If the problem were to find the solutions in the interval $\left[ -2\pi ,2\pi \right]$, then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,2\pi \right] :$

$$\begin{array}{lll} x&\approx& -5.384532,\ -4.0402457,\ -2.580... ...\ &&0.898653,\ 2.2429396,\ 3.70313096,\ 5.721646697 \end{array}$$

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

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