SOLVING TRIGONOMETRIC EQUATIONS

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


Problem 9.5d: Solve for x in the equation

\begin{displaymath}18\sin x\cos x=5\sin x\end{displaymath}

Answer:     The exact answers are

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&0\pm 2n\pi \\
&& \\
x_{2} &=&...
...ft( \displaystyle \frac{5}{18}\right) \pm 2n\pi \\
\end{array}\end{displaymath}

where n is an integer.

The approximate values of these solutions are

\begin{displaymath}\begin{array}{rclll}
x_{1} &\approx &0\pm 6.2831853n \\
x_{2...
...53n \\
x_{4} &\approx &-1.289316\pm 6.2831853n \\
\end{array}\end{displaymath}

where n is an integer.

Solution:

There are an infinite number of solutions to this problem. To solve for x, set the equation equal to zero and factor.


\begin{displaymath}\begin{array}{rclll}
18\sin x\cos x &=&5\sin x \\
&& \\
18\...
... \\
&& \\
\sin x\left( 18\cos x-5\right) &=&0 \\
\end{array}\end{displaymath}

then

\begin{displaymath}\begin{array}{rclll}
\sin x &=&0 \\
or&& \\
18\cos x-5 &=&0 \\
&& \\
\cos x &=&\displaystyle \frac{5}{18} \\
\end{array}\end{displaymath}

How do we isolate the x in the equations? We could take the arccosine of both sides of the cosine equation and the arcsine of both sides of the sine equations. However, the sine and cosine functions are not one-to-one functions.

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 each cosine equation. The graph of the sine 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 the sine equation.

\begin{displaymath}\begin{array}{rclll}
\sin x &=&0 \\
&& \\
\sin ^{-1}\left( ...
...t) \\
&& \\
x &=&\sin ^{-1}\left( 0\right) =0 \\
\end{array}\end{displaymath}

Since $\sin \left( \pi -x\right) =\sin \left( x\right) ,$ we know that if $\sin x=0,$ then $\sin \left( \pi -x\right) =0$ and

\begin{displaymath}\begin{array}{rclll}
\sin \left( \pi -x\right) &=&0 \\
&& \\...
...n ^{-1}\left( 0\right) =\pi -0\approx 3.1415927 \\
\end{array}\end{displaymath}

Now solve the equation $\cos x=\displaystyle \displaystyle \frac{5}{18}.$

\begin{displaymath}\begin{array}{rclll}
\cos x &=&\displaystyle \displaystyle \f...
...splaystyle \frac{5}{18}\right) \approx 1.289316 \\
\end{array}\end{displaymath}

We know that $\cos \left( -x\right) =\cos x.$ Therefore, if $\cos x=\displaystyle \displaystyle \frac{5}{18},$ then $\cos \left( -x\right) =\displaystyle \displaystyle \frac{5}{18}.$ Solve the equation $\cos \left( -x\right) =\displaystyle \displaystyle \frac{5}{18}.$

\begin{displaymath}\begin{array}{rclll}
\cos \left( -x\right) &=&\displaystyle \...
...playstyle \frac{5}{18}\right) \approx -1.289316 \\
\end{array}\end{displaymath}

Since the period of both the cosine function and the sine function is $2\pi $, the exact solutions are

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&0\pm 2n\pi \\
&& \\
x_{2} &=&...
...ft( \displaystyle \frac{5}{18}\right) \pm 2n\pi \\
\end{array}\end{displaymath}

where n is an integer.

The approximate values of these solutions are

\begin{displaymath}\begin{array}{rclll}
x_{1} &\approx &0\pm 6.2831853n \\
x_{2...
...53n \\
x_{4} &\approx &-1.289316\pm 6.2831853n \\
\end{array}\end{displaymath}

where n is an integer.

These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.

Numerical Check:

Check answer $x=\sin ^{-1}\left( 0\right) =0$

Left Side:

\begin{displaymath}18\sin x\cos x\approx 18\sin \left( 0\right) \cos \left(
0\right) \approx 0 \end{displaymath}

Right Side:         $5\sin x\approx 5\sin \left( 0\right) \approx 0 $

Since the left side equals the right side when you substitute 0 for x, then 0 is a solution.

Check answer $x=\pi -\sin ^{-1}\left( 0\right) =\pi \approx 3.1415927$

Left Side:

\begin{displaymath}18\sin x\cos x\approx 18\sin \left( 3.1415927\right) \cos
\left( 3.1415927\right) \approx 0 \end{displaymath}

Right Side:         $5\sin x\approx 5\sin \left( 3.1415927\right) \approx
0 $

Since the left side equals the right side when you substitute 3.1415927for x, then 3.1415927 is a solution.

Check answer $x=\cos ^{-1}\left( \displaystyle \displaystyle \frac{5}{18}\right) \approx 1.289316$

Left Side:

\begin{displaymath}18\sin x\cos x\approx 18\sin \left( 1.289316\right) \cos
\left( 1.289316\right) \approx 4.80323 \end{displaymath}

Right Side:         $5\sin x\approx 5\sin \left( 1.289316\right) \approx
4.80323 $

Since the left side equals the right side when you substitute 1.289316 for x, then 1.289316 is a solution.

Check answer $x=-\cos ^{-1}\left( \displaystyle \displaystyle \frac{5}{18}\right) \approx -1.289316$

Left Side:

\begin{displaymath}18\sin x\cos x\approx 18\sin \left( -1.289316\right) \cos
\left( -1.289316\right) \approx -4.80323 \end{displaymath}

Right Side:         $5\sin x\approx 5\sin \left( -1.289316\right) \approx
-4.80323 $

Since the left side equals the right side when you substitute -1.289316for x, then -1.289316 is a solution.

Graphical Check:

Graph the equation $f(x)=18\sin x\cos x-5\sin x.$ Note that the graph crosses the x-axis many times indicating many real solutions.

Verify that it crosses at 0. Since the period is $2\pi \approx 6.2831853$, it crosses again at 0+6.2831853=6.2831853 and at 0-6.2831853=-6.2831853, etc.

Verify that it crosses at 3.1415927. Since the period is $2\pi \approx 6.2831853$, it crosses again at 3.1415927+6.2831853=9.42477796 and at 3.1415927-6.2831853=-3.14159265, etc.

Verify that it crosses at 1.289316. Since the period is $2\pi \approx 6.2831853$, it crosses again at 1.289316+6.2831853=7.5725 and at 1.289316-6.2831853=-4.9938693, etc.

Verify that it crosses at -1.289316. Since the period is $2\pi \approx 6.2831853$, it crosses again at -1.289316+6.2831853=4.9938693 and at -1.289316-6.2831853=-7.57250, etc.

Note: If the problem were to find the solutions in the interval $\left[
0,2\pi \right] $, then you would choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,2\pi \right] :$ $x\approx 0,\ 1.289316,\ 3.1415927$, and 4.9938693.



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Author: Nancy Marcus

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