SOLVING TRIGONOMETRIC EQUATIONS

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


Example 1:        Solve for x in the following equation.



\begin{displaymath}2\sin^ {2}\left( x\right) +3\cos \left( x\right) -3=0\end{displaymath}


There are an infinite number of solutions to this problem.



We can make the solution easier if we convert all the trigonometric terms to cosine.


One common trigonometric identity is $\sin ^{2}(x)+\cos ^{2}(x)=1$ and $\sin
^{2}\left( x\right) =1-\cos ^{2}\left( x\right) \bigskip $ Replace the $%
\sin^ {2}\left( x\right) $. in the original equation with $1-\cos ^{2}\left(
x\right) $ and we have an equivalent equation with cosines terms.

\begin{displaymath}\begin{array}{lcl}
&& \\
2\sin^ {2}\left( x\right) +3\cos \l...
...eft( x\right) -3\cos \left( x\right) +1 &=&0 \\
&&
\end{array}\end{displaymath}


\begin{displaymath}\begin{array}{rclll}
&& \\
2\sin^ {2}\left( x\right) +3\cos ...
...eft( x\right) -3\cos \left( x\right) +1 &=&0 \\
&&
\end{array}\end{displaymath}

Isolate the cosine term. To do this, rewrite the left side of the equation in an equivalent factored form.


\begin{displaymath}\begin{array}{rclll}
&& \\
2\cos ^{2}\left( x\right) -3\cos ...
... (x)-1\right) \left( 2\cos (x)-1\right) &=&0 \\
&&
\end{array}\end{displaymath}

The product of two factors equals zero if at least one of the factors equals zero. This means that $\left( \cos (x)-1\right) \left( 2\cos (x)-1\right)
=0\ $if $\cos \left( x\right) -1=0$ or $2\cos \left( x\right) )-1=0.\bigskip
\bigskip $

We just transformed a difficult problem into two easier problems. To find the solutions to the original equation, $\left( \cos (x)-1\right) \left( 2\cos (x)-1\right)
=0\ $, we find the solutions to the equations $\cos (x)-1=0$and $2\cos (x)-1=0.\bigskip\bigskip $


\begin{displaymath}\begin{array}{rclll}
\cos (x)-1 &=&0 \\
&& \\
\cos \left( x\right) &=&1 \\
\end{array}\end{displaymath}

and

\begin{displaymath}\begin{array}{rclll}
2\cos (x)-1 &=&0 \\
&& \\
\cos \left( ...
...) &=&\displaystyle \displaystyle \frac{1}{2} \\
&&
\end{array}\end{displaymath}

How do we isolate the x? We could take the arccosine of both sides. 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 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 equation.


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

We know that $\cos \left( x\right) =\cos \left( -x\right) .$ Therefore, if $%
\cos (x)=0$, then $\cos (-x)=1.$


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

We complete the problem by solving for the second factor.

\begin{displaymath}\begin{array}{rclll}
&& \\
(3)\qquad \cos (x) &=&\displaysty...
...\displaystyle \frac{1}{2})\approx 1.04719755 \\
&&
\end{array}\end{displaymath}

As stated previously, we know that $\cos \left( x\right) =\cos \left( -x\right) .$ Therefore, if $\cos (x)=\displaystyle \displaystyle \frac{1}{2}$, then $\cos (-x)=\displaystyle \displaystyle \frac{1%
}{2}.$


\begin{displaymath}\begin{array}{rclll}
&& \\
\left( 4\right) \qquad \cos (-x) ...
...frac{1}{2}\right) \approx -1.04719755 \\
&& \\
&&
\end{array}\end{displaymath}

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

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&0\pm 2n\pi \\
&& \\
x_{2} &=&...
...e \displaystyle \frac{1}{2}\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_{3} &\approx &-1.04719755\pm 6.2831853n \\
&&
\end{array}\end{displaymath}

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=0


Left Side:

\begin{displaymath}2\sin^ {2}\left( x\right) +3\cos \left( x\right) -3\approx
2\sin^ {2}\left( 0\right) +3\cos \left( 0\right) -3\approx 0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

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




Check solution x=1.04719755


Left Side:

\begin{displaymath}2\sin^ {2}\left( x\right) +3\cos \left( x\right) -3\approx
2\...
...9755\right) +3\cos \left( 1.04719755\right) -3\approx 0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side of the original equation equals the right side of the original equation when you substitute 1.04719755 for x, then 1.04719755is a solution.




Check solution x=-1.04719755


Left Side:

\begin{displaymath}2\sin^ {2}\left( x\right) +3\cos \left( x\right) -3\approx
2\...
...755\right) +3\cos \left( -1.04719755\right) -3\approx 0\bigskip\end{displaymath}

Right Side:        $0\bigskip $

Since the left side of the original equation equals the right side of the original equation when you substitute -1.04719755 for x, then -1.04719755is a solution.




We have just verified algebraically that the exact solutions are x=0 and $%
x=\pm \cos ^{-1}\left( \displaystyle \displaystyle \frac{1}{2}\right) $ and these solutions repeat every $\pm 2\pi $ units. The approximate values of these solutions are $%
x\approx 0$ and $\pm 1.04719755$ and these solutions repeat every $\pm
6.2831853$ units.




Graphical Check:


Graph the equation $f(x)=2\sin^ {2}\left( x\right) +3\cos \left( x\right) -3.$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.


The graph crosses the x-axis at 0. Since the period is $2\pi \approx
6.2831853$, you can verify that the graph also crosses the x-axis again at 6.2831853 and at $2\left( 6.2831853\right) $, etc.


The graph crosses the x-axis at $\ 1.04719755$ Since the period is $2\pi \approx
6.2831853$, the graph also crosses the x-axis again at 1.04719755+6.2831853=7.3303829 and at $1.04719755+2\left( 6.2831853\right)
=13.613568$, etc.$.\bigskip $

The graph crosses the x-axis at $\ -1.04719755$. Since the period is $2\pi \approx
6.2831853$, the graph also crosses the x-axis again at -1.04719755+6.2831853=5.235988 and at $-1.04719755+2\left( 6.2831853\right)
=11.519173$, 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 $\left[ 0,2\pi \right] :$ $x\approx 0,\quad
1.04719755$, 5.235988, and $6.2831853.\bigskip\bigskip\bigskip\bigskip $



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.


If you would like to go to the next section, click on Next.


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Author: Nancy Marcus

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