SOLVING TRIGONOMETRIC EQUATIONS

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


Example 3:        Solve for x in the following equation.

\begin{displaymath}5\sin x\tan ^{2}x=\sin x\end{displaymath}

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}
5\sin x\tan ^{2}x &=&\sin x \\
&& \\
5...
...
&& \\
\sin x\left( 5\tan ^{2}x-1\right) &=&0 \\
\end{array}\end{displaymath}

then

\begin{displaymath}\begin{array}{rclll}
\sin x &=&0 \\
or && \\
5\tan ^{2}x-1 ...
...
\tan x &=&\pm \displaystyle \frac{1}{\sqrt{5}} \\
\end{array}\end{displaymath}

We know that $\sin x=\sin \left( \pi -x\right) .$ Therefore, if $\sin x=0,$then $\sin \left( \pi -x\right) =0. $

To solve for x, we have to isolate x. How do we isolate the x? We could take the inverse of both sides of the equations, arcsine in the sine equations and arctangent in the tangent equations. However, inverse functions can only be applied to one-to-one functions and neither the sine function nor the tangent function are one-to-one.

Let's restrict the domain so the functions so that they are one-to-one on the restricted domain while preserving the original range. The sine function is one-to-one on the interval $\left[ -\displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{\pi }{2}%
\right] $ and 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 both functions to the $\left( -\displaystyle \displaystyle \frac{\pi }{2},\displaystyle \displaystyle \frac{\pi }{2}\right) ,$we can take the inverses on this interval.


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


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


\begin{displaymath}\begin{array}{rclll}
(3)\qquad \tan x &=&\displaystyle \frac{...
...e \frac{1}{\sqrt{5}}\right) \approx 0.420534335 \\
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
(4)\qquad \tan x &=&-\displaystyle \frac...
... \frac{1}{\sqrt{5}}\right) \approx -0.420534335 \\
\end{array}\end{displaymath}

Since the period of the sine function is $2\pi $, therefore the solutions 0and $\pi $ repeat every $2\pi $ units. The period of the tangent function is $\pi $, therefore the solutions $\tan ^{-1}\left( \pm \displaystyle \displaystyle \frac{1}{\sqrt{5}}\right) $ repeat every $\pi $ units.

The exact solutions are

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&0\pm 2n\pi \\
&& \\
x_{2} &=&...
...playstyle \frac{1}{\sqrt{5}}\right) \pm n\pi \\
&&
\end{array}\end{displaymath}

where n is an integer.

The approximate values of theses solutions are

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&0\pm 6.2831853n \\
&& \\
x_{2...
...&& \\
x_{4} &=&-0.420534335\pm 3.141592653n \\
&&
\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=0

Left Side:

\begin{displaymath}5\sin x\tan ^{2}x\approx 5\sin \left( 0\right) \tan
^{2}\left( 0\right) =0 \end{displaymath}

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

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

Check answer x=3.141592653

Left Side:

\begin{displaymath}5\sin x\tan ^{2}x\approx 5\sin \left( 3.141592653\right)
\tan ^{2}\left( 3.141592653\right) =0 \end{displaymath}

Right Side:         $\sin x\approx \sin \left( 3.141592653\right) =0 $

Since the left side equals the right side when you substitute 3.141592653for x, then 3.141592653 is a solution.

Check answer x=0.420534335

Left Side:

\begin{displaymath}5\sin x\tan ^{2}x\approx 5\sin \left( 0.420534335\right)
\tan ^{2}\left( 0.420534335\right) \approx 0.40824829 \end{displaymath}

Right Side:         $\sin x\approx \sin \left( 0.420534335\right) \approx
0.40824829 $

Since the left side equals the right side when you substitute 0.420534335for x, then 0.420534335 is a solution.

Check answer x=-0.420534335

Left Side:

\begin{displaymath}5\sin x\tan ^{2}x\approx 5\sin \left( -0.420534335\right)
\tan ^{2}\left( -0.420534335\right) \approx 0.40824829 \end{displaymath}

Right Side:         $\sin x\approx \sin \left( -0.420534335\right) \approx
0.40824829 $

Since the left side equals the right side when you substitute -0.420534335for x, then -0.420534335 is a solution.

Graphical Check:

Graph the equation $f(x)=5\sin x\tan ^{2}x-\sin x$, formed by subtracting the right side of the original equation from the left side of the original equation. Note that the graph crosses the x-axis many times indicating many solutions. Verify that the graph crosses the x-axis at the points 0, $%
\pi ,\pm 0.420534335.$



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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Author: Nancy Marcus

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