SOLVING TRIGONOMETRIC EQUATIONS

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


Problem 9.6a: Solve for x in the equation

\begin{displaymath}6\sin ^{2}x+\sin x-1=0\end{displaymath}

Answer:    The exact answers are

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&\sin ^{-1}\left( \displaystyle ...
...ft( -\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 &0.3398369\pm 6.2831853n ...
...53n \\
x_{4} &\approx &3.6651914\pm 6.2831853n \\
\end{array}\end{displaymath}

where n is an integer.

Solution:

There are an infinite number of solutions to this problem.

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


\begin{displaymath}\begin{array}{rclll}
6\sin ^{2}x+\sin x-1 &=&0 \\
&& \\
\left( 3\sin x-1\right) \left( 2\sin x+1\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 zeros. This means that $6\sin ^{2}x+\sin x-1=0$ if $3\sin x-1=0\ $or $2\sin x+1=0.$

We have transformed a difficult problem into two problems. To find the solutions to the original equation, $\left( 3\sin x-1\right) \left( 2\sin x+1\right) =0$, we find the solutions to the equations $3\sin x-1=0\ $ and $2\sin x+1=0.$


\begin{displaymath}\begin{array}{rclll}
3\sin (x)-1 &=&0 \\
&& \\
\sin \left( x\right) &=&\displaystyle \frac{1}{3} \\
\end{array}\end{displaymath}

and

\begin{displaymath}\begin{array}{rclll}
2\sin (x)+1 &=&0 \\
&& \\
\sin (x) &=&-\displaystyle \frac{1}{2} \\
\end{array}\end{displaymath}

How do we isolate the x? We could take the arcsine of both sides. However, the sine function is not a one-to-one function.

We can restrict the domain of the function so that the function is one-to-one on the restricted domain while preserving the original range. 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 each equation.


\begin{displaymath}\begin{array}{rclll}
(1)\qquad \sin \left( x\right) &=&\displ...
...splaystyle \frac{1}{3}\right) \approx 0.3398369 \\
\end{array}\end{displaymath}

We know that $\sin \left( x\right) =\sin \left( \pi -x\right) .$ Therefore, if $\sin (x)=\displaystyle \displaystyle \frac{1}{3}$, then $\sin (\pi -x)=\displaystyle \displaystyle \frac{1}{3}.$


\begin{displaymath}\begin{array}{rclll}
\left( 2\right) \qquad \sin (\pi -x) &=&...
...splaystyle \frac{1}{3}\right) \approx 2.8017557 \\
\end{array}\end{displaymath}

We complete the problem by solving for the second factor.

\begin{displaymath}\begin{array}{rclll}
(3)\qquad \sin (x) &=&-\displaystyle \fr...
...-\displaystyle \frac{1}{2})\approx -0.523598776 \\
\end{array}\end{displaymath}

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


\begin{displaymath}\begin{array}{rclll}
\left( 4\right) \qquad \sin (\pi -x) &=&...
...splaystyle \frac{1}{2}\right) \approx 3.6651914 \\
\end{array}\end{displaymath}

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

\begin{displaymath}\begin{array}{rclll}
x_{1} &=&\sin ^{-1}\left( \displaystyle ...
...ft( -\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 &0.3398369\pm 6.2831853n ...
...
&& \\
x_{4} &\approx &3.6651914\pm 6.2831853n \\
\end{array}\end{displaymath}

where n is an integer.

One 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.

One 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 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 $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{1}{3}\right) \approx 0.3398369$

Left Side:

\begin{displaymath}6\sin ^{2}\left( x\right) +\sin \left( x\right) -1\approx 6\s...
...ft( 0.3398369\right) +\sin \left( 0.3398369\right) -1\approx 0 \end{displaymath}

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute 0.3398369 for x, then 0.3398369 is a solution.

Check solution $x=\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{1}{3}\right) \approx 2.8017557 $

Left Side:

\begin{displaymath}6\sin ^{2}\left( x\right) +\sin \left( x\right) -1\approx 6\s...
...ft( 2.8017557\right) +\sin \left( 2.8017557\right) -1\approx 0 \end{displaymath}

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute 2.8017557 for x, then 2.8017557 is a solution.

Check solution $x=\sin ^{-1}\left( -\displaystyle \displaystyle \frac{1}{2}\right) \approx -0.523598776$

Left Side:

\begin{displaymath}6\sin ^{2}\left( x\right) +\sin \left( x\right) -1\approx
6\s...
....523598776\right) +\sin \left( -0.523598776\right)
-1\approx 0 \end{displaymath}

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute -0.523598776 for x, then <tex2htmlcommentmark> -0.523598776 is a solution.

Check solution $x=\pi -\sin ^{-1}\left( -\displaystyle \displaystyle \frac{1}{2}\right) \approx 3.6651914$

Left Side:

\begin{displaymath}6\sin ^{2}\left( x\right) +\sin \left( x\right) -1\approx 6\s...
...ft( 3.6651914\right) +\sin \left( 3.6651914\right) -1\approx 0 \end{displaymath}

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute 3.6651914 for x, then 3.6651914 is a solution.

We have just verified that the exact solutions $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{1}{3}\right) \pm 2n\pi ,\ ...
... \sin ^{-1}\left( -\displaystyle \displaystyle \frac{1}{2}\right) \pm 2n\pi ,\ $ and $\pi -\sin^{-1}\left( -\displaystyle \displaystyle \frac{1}{2}\right) \pm 2n\pi ,\ $ are the exact solutions and these solutions repeat every $\pm 2\pi $ units. The approximate values of these solutions are $x\approx 0.3398369,\ 2.8017557,$ -0.523598776, and 3.6651914 and these solutions repeat every $\pm 6.2831853$ units.

Graphical Check:

Graph the equation $f(x)=6\sin ^{2}\left( x\right) +\sin \left( x\right) -1.$Note that the graph crosses the x-axis many times indicating many solutions.


The graph crosses the x-axis at 0.3398369. Since the period is $2\pi
\approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 0.3398369+6.2831853=6.6230222 and at $0.3398369+2\left( 6.2831853\right) =12.9062075$, etc.

The graph crosses the x-axis at $\ 2.8017557$. Since the period is $2\pi
\approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 2.8017557+6.2831853=9.084941007 and at $2.8017557+2\left( 6.2831853\right) =15.3681263$, etc.

The graph crosses the x-axis at $\ -0.523598776$. Since the period is $2\pi
\approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> -0.523598776+6.2831853=5.7595865 and at $-0.523598776+2\left(
6.2831853\right) =12.04277184$, etc.

The graph crosses the x-axis at $\ 3.6651914$. Since the period is $2\pi
\approx 6.2831853$, the graph also crosses the x-axis again at <tex2htmlcommentmark> 3.6651914+6.2831853=9.9483767 and at $3.6651914+2\left( 6.2831853\right) =16.231562$, 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 $\left[ 0,2\pi \right] :$ $x\approx
0.3398369,\quad 2.8017557$, 5.7595865 and 3.6651914.



If you would like to review problem 9.4b, click on problem 9.4b.

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Author: Nancy Marcus

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