SOLVING TRIGONOMETRIC EQUATIONS

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


Problem 9.7b:        Solve for x in the equation

\begin{displaymath}6\cos ^{2}x+13\sin x+11=0\end{displaymath}



Answer:    The exact answers are

\begin{displaymath}\begin{array}{rclll}
&& \\
x_{1} &=&\sin ^{-1}\left( \displa...
...}\left( \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 &0.52359878\pm 6.2...
... \\
&& \\
x_{2} &\approx &2.6179939\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 $6\cos ^{2}x+13\sin x+11=0,$ there are two different trigonometric terms. The problem will be less difficult to solve if you convert the equation to an equivalent equation with just sine terms or just cosine terms.


Which one shall it be? The easiest way is to convert the trigonometric terms to sine terms because from the identity $\sin ^{2}\left( x\right) +\cos
^{2}\left( x\right) =1$, you can solve for $\cos ^{2}\left( x\right) =1-\sin
^{2}\left( x\right) $. Substitute $1-\sin ^{2}\left( x\right) $ for $\cos
^{2}\left( x\right) $ in the original equation.


\begin{displaymath}\begin{array}{rclll}
&& \\
6\cos ^{2}x+13\sin x+11 &=&0 \\
...
...&=&0 \\
&& \\
6\sin ^{2}x-13\sin x-17 &=&0 \\
&&
\end{array}\end{displaymath}

Rewrite the equation $6\sin ^{2}x-13\sin x-17=0$ in factored form and solve.

\begin{displaymath}\begin{array}{rclll}
&& \\
6\sin ^{2}x-13\sin x-17 &=&0 \\
...
...\sin x-5\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 $\left( 3\sin x-5\right) \left(
2\sin x-1\right) =0$ if $3\sin x-5=0\ $or $2\sin x-1=0.\bigskip\bigskip $

We have transformed a difficult problem into two less difficult problems. To find the solutions to the original equation, $\left( 3\sin x-5\right) \left(
2\sin x-1\right) =0$, we first solve for $\sin \left( x\right) $ in the in the equations $3\sin x-5=0\ $ and $2\sin x-1=0.\bigskip $


\begin{displaymath}\begin{array}{rclll}
3\sin x-5 &=&0 \\
&& \\
\sin \left( x\...
... &=&\displaystyle \frac{5}{3}>1\rightarrow \phi \\
\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}

To solve for x, we must first isolate the x, and just 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 could restrict the domain of the function so that 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) &=&\displ...
...ystyle \frac{1}{2}\right) \approx 0.52359878 \\
&&
\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 $\sin \left( \pi -x\right) =\displaystyle \displaystyle \frac{1}{2}.$


\begin{displaymath}\begin{array}{rclll}
&& \\
\left( 2\right) \qquad \sin (\pi ...
... \frac{1}{2}\right) \approx 2.6179939 \\
&& \\
&&
\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 ...
...}\left( \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 &0.52359878\pm 6.2...
...\\
&& \\
x_{2} &\approx &2.61799388\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 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 $x=\sin ^{-1}\left( \displaystyle \displaystyle \frac{1}{2}\right) \approx 0.52359878$


Left Side:

\begin{displaymath}6\cos ^{2}x+13\sin x+11
\approx 6\cos ^{2}\left( 0.52359878\right) +13\sin \left( 0.52359878\right)
+11\approx 0 \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 0.52359878 for x, then 0.52359878is a solution.




Check solution $x=\pi -\sin ^{-1}\left( \displaystyle \displaystyle \frac{1}{2}\right) \approx
2.6179939 $


Left Side:

\begin{displaymath}6\cos ^{2}x+13\sin x+11
\approx 6\cos ^{2}\left( 2.6179939\right) +13\sin \left( 2.6179939\right)
+11\approx 0\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 2.6179939 for x, then 2.6179939 is a solution.




The exact solutions are $x=\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 ,\ $ and these solutions repeat every $\pm 2\pi $ units. The approximate values of these solutions are $x\approx 0.52359878$ and 2.6179939 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 0.52359878. Since the period is $%
2\pi \approx 6.2831853$, the graph also crosses the x-axis again at 0.52359878+6.2831853=6.806784 and at $0.52359878+2\left( 6.2831853\right)
=13.089969$, etc.


Verify the graph crosses the x-axis at $\ 2.6179939$. Since the period is $%
2\pi \approx 6.2831853$, the graph also crosses the x-axis again at 2.6179939+6.2831853=8.901179 and at $2.6179939+2\left( 6.2831853\right)
=15.1843645$, 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 $\left[ 0,2\pi \right] :$ $x\approx 0.52359878$ and $\ 2.6179939.\bigskip\bigskip\bigskip\bigskip $

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Author: Nancy Marcus

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