SOLVING TRIGONOMETRIC EQUATIONS


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


Problem 9.7a:         Solve for x in the equation

\begin{displaymath}35\sin ^{2}x+26\cos x-38=0\end{displaymath}


Answer:    The exact answers are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &=& \cos ^{-1}\left( \displ...
...}\left( \displaystyle \frac{3}{5}\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 &1.4274488\pm 6.28...
...\\
&& \\
x_{4} &\approx &-0.9272952\pm 6.2831853n
\end{array}\end{displaymath}

$\quad $where n is an integer.




Solution:


There are an infinite number of solutions to this problem.



In the equation $35\sin ^{2}x+26\cos x-38=0,$ there are two different trigonometric terms. To make life easier for yourself, convert the equation to an equivalent equation in terms of just sine terms or just cosine terms. Which one shall it be? The easiest way is to convert the trigonometric terms to cosine terms because from the identity $\sin ^{2}\left( x\right) +\cos
^{2}\left( x\right) =1$, you can solve for $\sin ^{2}\left( x\right) =1-\cos
^{2}\left( x\right) $. Substitute $1-\cos ^{2}\left( x\right) $ for $\sin
^{2}\left( x\right) $ in the original equation.


\begin{displaymath}\begin{array}{rclll}
&& \\
35\sin ^{2}x+26\cos x-38 &=&0 \\ ...
...&=&0 \\
&& \\
35\cos ^{2}x-26\cos x+3 &=&0 \\
&&
\end{array}\end{displaymath}

Rewrite the equation $35\cos ^{2}x-26\cos x+3=0$ in factored form and solve for $\cos (x)$.

\begin{displaymath}\begin{array}{rclll}
&& \\
35\cos ^{2}x-26\cos x+3 &=&0 \\
...
...\cos x-1\right) \left( 5\cos x-3\right) &=&0 \\
&&
\end{array}\end{displaymath}

The only way the product of two factors equals zero is if at least one of the factors equals zero. This means that $35\cos ^{2}x-26\cos x+3=0$ if $%
7\cos x-1=0\ $or $5\cos x-3=0.\bigskip\bigskip\bigskip $

We have transformed a difficult problem into two less difficult problems. To find the solutions to the original equation, $\left( 7\cos x-1\right) \left(
5\cos x-3\right) =0$, we first find the solutions to $\cos (x)$ in the equations $7\cos x-1=0$ and $5\cos x-3=0.$


\begin{displaymath}\begin{array}{rclll}
&& \\
7\cos x-1 &=&0 \\
&& \\
\cos \left( x\right) &=&\displaystyle \frac{1}{7} \\
\end{array}\end{displaymath}

and

\begin{displaymath}\begin{array}{rclll}
5\cos (x)-3 &=&0 \\
&& \\
\cos (x) &=&\displaystyle \frac{3}{5} \\
&&
\end{array}\end{displaymath}

To solve for x, we must isolate x, and just how do we do that? We could take the arccosine of both sides. However, the cosine function is not a one-to-one function.


We can restrict the domain of the function so the function is one-to-one on the restricted domain while preserving the original range. 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) &=&\displ...
...aystyle \frac{1}{7}\right) \approx 1.4274488 \\
&&
\end{array}\end{displaymath}

We know that $\cos \left( x\right) =\cos \left( -x\right) .$ Therefore, if $%
\cos (x)=\displaystyle \displaystyle \frac{1}{7}$, then $\cos \left( -x\right) =\displaystyle \displaystyle \frac{1}{7}.$


\begin{displaymath}\begin{array}{rclll}
&& \\
\left( 2\right) \qquad \cos (-x) ...
...\frac{1}{7}\right) \approx -1.4274488 \\
&& \\
&&
\end{array}\end{displaymath}

We complete the problem by solving for the second factor.

\begin{displaymath}\begin{array}{rclll}
&& \\
(3)\qquad \cos (x) &=&\displaysty...
...laystyle \frac{3}{5})\approx 0.927295 \\
&& \\
&&
\end{array}\end{displaymath}

We know that $\cos \left( x\right) =\cos \left( -x\right) .$ Therefore, if $%
\cos (x)=\displaystyle \displaystyle \frac{3}{5}$, then $\cos \left( -x\right) =\displaystyle \displaystyle \frac{3}{5}.$


\begin{displaymath}\begin{array}{rclll}
&& \\
\left( 2\right) \qquad \cos (-x) ...
... \frac{3}{5}\right) \approx -0.927295 \\
&& \\
&&
\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} &=& \cos ^{-1}\left( \displaystyle...
...}\left( \displaystyle \frac{3}{5}\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 &1.4274488\pm 6.28...
... \\
&& \\
x_{4} &\approx &-0.927295\pm 6.2831853n
\end{array}\end{displaymath}

where n is an integer.




You can check each solution algebraically by substituting the 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 the left side of the original equation and graphing the function formed by the right side of the original equation. The x-coordinates of the points of intersection are the solutions. The right side of the equation is 0 and <tex2htmlcommentmark> f(x)=0 is the x-axis. So really what you are looking for are the x-intercepts to the function formed by the left side of the equation.


Algebraic Check:


Check solution $x= \cos ^{-1}\left( \displaystyle \displaystyle \frac{1}{7}\right) \approx 1.4274488$


Left Side:

\begin{displaymath}\begin{array}{rclll}
35\sin ^{2}x+26\cos x-38
&\approx& 35\si...
... +26\cos \left( 1.4274488\right) -38\\
&\approx& 0
\end{array}\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 1.4274488 for x, then 1.4274488 is a solution.




Check solution $x=- \cos ^{-1}\left( \displaystyle \displaystyle \frac{1}{7}\right) \approx -1.4274488$


Left Side:

\begin{displaymath}\begin{array}{rclll}
35\sin ^{2}x+26\cos x-38
&\approx &35\s...
...+26\cos \left(
-1.4274488\right) -38\\
&\approx& 0
\end{array}\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 -1.4274488 for x, then -1.4274488is a solution.




Check solution $x= \cos ^{-1}\left( \displaystyle \displaystyle \frac{3}{5}\right) \approx 0.927295$


Left Side:

\begin{displaymath}\begin{array}{rclll}
35\sin ^{2}x+26\cos x-38
&\approx &35\s...
...) +26\cos \left( 0.927295\right)
-38\\
&\approx& 0
\end{array}\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 0.927295 for x, then 0.927295 is a solution.




Check solution $x=- \cos ^{-1}\left( \displaystyle \displaystyle \frac{3}{5}\right) \approx -0.927295$


Left Side:

\begin{displaymath}\begin{array}{rclll}
35\sin ^{2}x+26\cos x-38
&\approx &35\s...
...) +26\cos \left( -0.927295\right)-38\\
&\approx& 0
\end{array}\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 -0.927295 for x, then -0.927295 is a solution. We have just verified that the exact solutions are $x=\pm \cos
^{-1}\left( \displaystyle \displaystyle \frac{1}{7}\right) \pm 2n\pi \ $and $\pm \cos ^{-1}\left(
\displaystyle \displaystyle \frac{3}{5}\right) \pm 2n\pi ,\ $ and these solutions repeat every $\pm
2\pi $ units. The approximate values of these solutions are $x\approx \pm
1.4274488$ and $\pm 0.927295$ and these solutions repeat every $\pm 6.2831853
$ units.




Graphical Check:


Graph the equation $f(x)=35\sin ^{2}x+26\cos x-38.$ Note that the graph crosses the x-axis many times indicating many solutions.


Verify the graph crosses the x-axis at -0.927295. Since the period is $%
2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> -0.927295+6.2831853=5.35589 and at $-0.927295+2\left( 6.2831853\right)
=11.6390756$, etc.


Verify the graph crosses the x-axis at $\ 0.927295$. Since the period is $%
2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 0.927295+6.2831853=7.21048 and at $2.8017557+2\left( 6.2831853\right)
=13.493666$, etc.

Verify the graph crosses the x-axis at $\ 1.4274488$. Since the period is $%
2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 1.4274488+6.2831853=7.710634 and at $1.4274488+2\left( 6.2831853\right)
=13.993819$, etc.


Verify the graph crosses the x-axis at $\ -1.4274488$. Since the period is $%
2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> -1.4274488+6.2831853=4.8557365 and at $-1.4274488+2\left( 6.2831853\right)
=11.1389218$, 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 $\left[ 0,2\pi \right] :$ $x\approx
0.927295,\ 1.4274488,\ 4.8557365,$ and $5.35589.\bigskip\bigskip\bigskip
\bigskip $

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Author: Nancy Marcus

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