SOLVING TRIGONOMETRIC EQUATIONS

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


Example 2:        Solve for x in the following equation. $2\tan x\sin ^{2}x=\tan x$

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

then

\begin{displaymath}\begin{array}{rclll}
\tan x &=&0 \\
&& \\
\displaystyle \frac{\sin x}{\cos x} &=&0 \\
&& \\
\sin x &=&0 \\
\end{array}\end{displaymath}

or

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

We know that $\sin x=\sin \left( \pi -x\right) .$ Therefore, if $\sin x=\pm
\sqrt{\displaystyle \displaystyle \frac{1}{2}},$ then $\sin \left( \pi -x\right) =\pi -\pm \sqrt{\displaystyle \displaystyle \frac{%
1}{2}}$ and if $\sin x=0$, then $\sin \left( \pi -x\right) =0 $

To solve for x, we must 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 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 each equation.


\begin{displaymath}\begin{array}{rclll}
(1)\qquad \sin \left( x\right) &=&\sqrt{...
...playstyle \frac{1}{2}}\right) \approx 0.7853982 \\
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
(2)\qquad \sin \left( \pi -x\right) &=&\...
...playstyle \frac{1}{2}}\right) \approx 2.3561945 \\
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
(3)\qquad \sin \left( x\right) &=&-\sqrt...
...laystyle \frac{1}{2}}\right) \approx -0.7853982 \\
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
(4)\qquad \sin \left( \pi -x\right) &=&-...
...splaystyle \frac{1}{2}}\right) \approx 3.926991 \\
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
(5)\qquad \sin \left( x\right) &=&0 \\
...
...\
&& \\
x_{5} &=&\sin ^{-1}\left( 0\right) =0 \\
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
(6)\qquad \sin \left( \pi -x\right) &=&0...
... -\sin ^{-1}\left( 0\right) \approx 3.141592653 \\
\end{array}\end{displaymath}

Since the period of the sine function is $2\pi $, these solutions repeat every $2\pi $ units. The exact solutions are $x=\sin ^{-1}\left( \pm \sqrt{%
\displaystyle \displaystyle \frac{1}{2}}\right) ,$ $x=\pi -\sin ^{-1}\left( \pm \sqrt{\displaystyle \displaystyle \frac{1}{2}}%
\right) $, x=0, $x=\pi ,$ and these solutions repeat every $2\pi n$radians, where n is an integer.

The approximate values of theses solutions are $x\approx \pm 0.7853982,$ 2.3561945, 3.926991, 0 and 3.14159265, and these solutions repeat every 6.2831853n radians 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 the answer x=-0.7853982

Left Side:

\begin{displaymath}2\tan x\sin ^{2}x\approx 2\tan \left( -0.7853982\right)
\sin ^{2}\left( -0.7853982\right) \approx -1 \end{displaymath}

Right Side:

\begin{displaymath}\tan x\approx \tan \left( -0.7853982\right) \approx
-1 \end{displaymath}

Since the left side equals the right side when you substitute -0.7853982for x, then -0.7853982 is a solution.

Check answer x=0.7853982

Left Side:

\begin{displaymath}2\tan x\sin ^{2}\left( 0.7853982\right) \approx 2\tan
x\sin ^{2}\left( 0.7853982\right) \approx 1 \end{displaymath}

Right Side:

\begin{displaymath}\tan x\approx \tan \left( 0.7853982\right) \approx
1 \end{displaymath}

Since the left side equals the right side when you substitute 0.7853982for x, then 0.7853982 is a solution.

Check answer x=2.3561945

Left Side:

\begin{displaymath}2\tan x\sin ^{2}\left( 2.3561945\right) \approx 2\tan
x\sin ^{2}\left( 2.3561945\right) \approx -1 \end{displaymath}

Right Side:

\begin{displaymath}\tan x\approx \tan \left( 2.3561945\right) \approx
-1 \end{displaymath}

Since the left side equals the right side when you substitute 2.3561945for x, then 2.3561945 is a solution.

Check answer x=3.926991

Left Side:

\begin{displaymath}2\tan x\sin ^{2}x\approx 2\tan \left( 3.926991\right)
\sin ^{2}\left( 3.926991\right) \approx 1 \end{displaymath}

Right Side:

\begin{displaymath}\tan x\approx \tan \left( 3.926991\right) \approx
1 \end{displaymath}

Since the left side equals the right side when you substitute 3.926991 for x, then 3.926991 is a solution.

Check answer x=0

Left Side:

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

Right Side:

\begin{displaymath}\tan x\approx \tan 0\approx 0 \end{displaymath}

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

Check answer x=3.1415925

Left Side:

\begin{displaymath}2\tan x\sin ^{2}x\approx 2\tan \left( 3.1415925\right)
\sin ^{2}\left( 3.1415925\right) \approx 0 \end{displaymath}

Right Side:

\begin{displaymath}\tan x\approx \tan \left( 3.1415925\right) \approx
0 \end{displaymath}

Since the left side equals the right side when you substitute 3.1415925for x, then 3.1415925 is a solution.

Graphical Check:

Graph the equation $f(x)=2\tan x\sin ^{2}x-\tan x$, 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.

Note that it crosses at 0.7853982. Since the period is $2\pi \approx
6.2831853$, it crosses again at 0.7853982+6.2831853=7.0685835 and at <tex2htmlcommentmark> 0.7853982+2(6.2831853)=13.351769, etc.

Note that it crosses at -0.7853982. Since the period is $2\pi \approx
6.2831853$, it crosses again at -0.7853982+6.2831853=5.497787 and at <tex2htmlcommentmark> -0.7853982+2(6.2831853)=11.78097, etc.

Note that it crosses at 2.3561945. Since the period is $2\pi \approx
6.2831853$, it crosses again at 2.3561945+6.2831853=8.6393798 and at <tex2htmlcommentmark> 2.3561945+2(6.2831853)=14.922565, etc.

Note that it crosses at 3.926991. Since the period is $2\pi \approx
6.2831853$, it crosses again at 3.926991+6.2831853=10.210176 and at <tex2htmlcommentmark> 3.926991+2(6.2831853)=16.49336, etc.

Note 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+2(6.2831853)=12.56637 , etc.

Note that it crosses at 3.141592653. Since the period is $2\pi \approx
6.2831853$, it crosses again at 3.141592653+6.2831853=9.424778 and at <tex2htmlcommentmark> 3.141592653+2(6.2831853)=15.707963, etc.



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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