SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.10d:        Solve for x in the equation

$$4\cot x+5\sin x=0$$

Answer:    The exact answers are

$$\begin{array}{rclll} && \\ x &=&\pm \cos ^{-1}\left( \displaystyle \frac{2-\sqrt{29}}{5}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x &\approx &\pm 2.31451988\pm 6.2831853n \\ && \end{array}$$

$\quad$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.

Rewrite the original equation $4\cot x+5\sin x=0\$ as $\ \displaystyle \displaystyle \frac{4\cos x}{% \sin x}+5\sin x=0.\bigskip$

Multiply both sides of the equation by $\sin x:\qquad 4\cos x+5\sin ^{2}x=0$

One common trigonometric identity is $\sin ^{2}x+\cos ^{2}x=1$ where $\sin ^{2}x=1-\cos ^{2}x.$ If we replace the $\sin ^{2}x$ term with $1-\cos ^{2}x$ , all the trigonometric terms will be cosine terms.

Replace $\sin ^{2}x$ with $1-\cos ^{2}x$ in the original equation and simplify.

$$\begin{array}{rclll} && \\ 4\cos x+5\sin ^{2}x &=&0 \\ && \... ...\\ && \\ 5\cos ^{2}x-4\cos x-5 &=&0 \\ && \\ && \end{array}$$

Isolate the cosine term. We do this by solving for $\cos x$ using the Quadratic Formula.

$$\begin{array}{rclll} && \\ 5\cos ^{2}x-4\cos x-5 &=&0 \\ &&... ...=&\displaystyle \frac{2-\sqrt{29}}{5} \\ && \\ && \end{array}$$

How do we isolate the x in each of this equation? We could take the inverse (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 tangent 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.

$$\begin{array}{rclll} \cos x &=&\displaystyle \frac{2-\sqrt{29... ...-\sqrt{29}}{5}\right) \approx 2.31452 \\ && \\ && \end{array}$$

We know that $\cos x=\cos \left( -x\right)$, therefore if $\cos x=\displaystyle \displaystyle \frac{2-% \sqrt{29}}{5}$, then $\cos \left( -x\right) =\displaystyle \displaystyle \frac{2-\sqrt{29}}{5}.$

$$\begin{array}{rclll} && \\ \cos \left( -x\right) &=&\display... ...\sqrt{29}}{5}\right) \approx -2.31452 \\ && \\ && \end{array}$$

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

$$\begin{array}{rclll} && \\ x &=&\pm \cos ^{-1}\left( \displaystyle \frac{2-\sqrt{29}}{5}\right) \pm 2n\pi \\ && \end{array}$$

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} && \\ x &\approx &\pm 2.31452\pm 6.2831853n \\ && \end{array}$$

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 $x= \cos ^{-1}\left( \displaystyle \displaystyle \frac{2-\sqrt{29}}{5}\right) \approx 2.31452\bigskip$

Left Side:

$$\begin{array}{rclll} \cot x+5\sin x &=&\displaystyle \frac{4}... ...ght) }+5\sin \left( 2.31452\right) \approx 0 \\ && \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute 2.31452 for x, then 2.31452 is a solution.

Check solution $x=- \cos ^{-1}\left( \displaystyle \displaystyle \frac{2-\sqrt{29}}{5}\right) \approx -2.31452\bigskip$

Left Side:

$$\begin{array}{rclll} 4\cot x+5\sin x &=&\displaystyle \frac{4... ...ht) }+5\sin \left( -2.31452\right) \approx 0 \\ && \end{array}$$

Right Side:        $0\bigskip$

Since the left side of the original equation equals the right side of the original equation when you substitute -2.31452 for x, then -2.31452 is a solution.

Graphical Check:

Graph the equation $f\left( x\right) =4\cot x+5\sin x\$or $\ f\left( x\right) =\displaystyle \frac{4}{\tan x}+5\sin x.$

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 2.31452. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at 2.31452+6.2831853=8.597705 and at $2.31452+2\left( 3.141592653\right) =14.88089$, etc.

Verify the graph crosses the x-axis at -2.31452. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at -2.31452+6.2831853=3.9686653 and at $-2.31452+2\left( 3.141592653\right) =10.25185$, 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 $\left[ 0,2\pi \right] :$ $x\approx 2.31452$and $3.9686653.\bigskip\bigskip\bigskip\bigskip$

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