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Wednesday, 30 July 2014

CHAPTER 8- SOLVING TRIGONOMETRIC EQUATIONS

This sections illustrates the process of solving trigonometric equations of various forms. It also shows you how to check your answer three different ways: algebraically, graphically, and using the concept of equivalence.The following table is a partial lists of typical equations

Solve for x in the following equation.


Example 1:        
$2sin\left( x\right) -1=0$

There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.



\begin{eqnarray*}&& \\
2\sin \left( x\right) -1 &=&0 \\
&& \\
\sin \left( x\right) &=&\displaystyle \frac{1}{2} \\
&& \\
&&
\end{eqnarray*}



We know that the $\sin \left( \displaystyle \frac{\pi }{6}\right) =\displaystyle \frac{1}{2},$therefore $x=\displaystyle \frac{\pi }{6}.$ The sine function is positive in quadrants I and II. The $\sin \left( \pi -\displaystyle \frac{\pi }{6}\right) =\sin \displaystyle \frac{5\pi }{6}$is also equal to $\displaystyle \frac{1}{2}.$ Therefore, two of the solutions to the problem are $x=\displaystyle \frac{\pi }{6}$and $x=\displaystyle \frac{5\pi }{6}.
$


The period of the sin $\left( x\right) $ function is $2\pi .$ This means that the values will repeat every $2\pi $ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{\pi }{6}\pm n\left( 2\pi \right) $ and $x=
\displaystyle \frac{5\pi }{6}\pm n\left( 2\pi \right) $ where n is an integer. The approximate solutions are $x=0.523598775598\pm n\left( 2\pi \right) $ and $
x=2.61799387799\pm n\left( 2\pi \right) $ 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 answer . $x=\displaystyle \frac{\pi }{6}$


  • Left Side: $\qquad 2\sin \left( x\right) -1=2\sin \left( \displaystyle \frac{\pi }{6}
\right) -1=2\left( \displaystyle \frac{1}{2}\right) -1=0 $
  • Right Side:        0

Since the left side equals the right side when you substitute $\displaystyle \frac{\pi }{6
}$ for x, then $\displaystyle \frac{\pi }{6
}$ is a solution.


Check answer . $x=\displaystyle \frac{5\pi }{6}$


  • Left Side: $\qquad 2\sin \left( x\right) -1=2\sin \left( \displaystyle \frac{5\pi }{6}
\right) -1=2\left( \displaystyle \frac{1}{2}\right) -1=0 $
  • Right Side:        0
Since the left side equals the right side when you substitute $\displaystyle \frac{5\pi }{
6}$ for x, then $\displaystyle \frac{5\pi }{
6}$ is a solution.


Graphical Check:
Graph the equation

f (x) = 2 sin(x) - 1

Note that the graph crosses the x-axis many times indicating many solutions. Note that it crosses at $\displaystyle \frac{
\pi }{6}\approx 0.5236$. Since the period is $2\pi \approx 6.2831$, it crosses again at 0.5236+6.283=6.81 and at0.5236+2(6.283)=13.09, etc. The graph crosses at $\displaystyle \frac{5\pi }{6}\approx 2.618$. Since the period is $
2\pi \approx 6.283$, it will cross again at 2.618+6.283=8.9011 and at 2.618+2(6.283)=15.18, etc.



Example 2:        Solve for x in the following equation.

\begin{displaymath}2\sin^{2}\left( x\right) -\sin \left( x\right) -1=0\end{displaymath}



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



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

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



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


and

\begin{displaymath}\begin{array}{rclll}
\sin (x)-1 &=&0 \\
&& \\
\sin (x) &=&1 \\
\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.

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 sine function is one-to-one on the interval $\left[ -\displaystyle \frac{\pi }{2}, \displaystyle \frac{\pi }{2}\right] .$ 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) &=&-\disp...
...aystyle \frac{1}{2}\right) \approx -0.523598776 \\
\end{array}\end{displaymath}



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



\begin{displaymath}\begin{array}{rclll}
\left( 2\right) \qquad \sin (\pi -x) &=&...
...isplaystyle \frac{1}{2}\right) \approx 3.665191 \\
\end{array}\end{displaymath}



We complete the problem by solving for the second factor.

\begin{displaymath}\begin{array}{rclll}
(3)\qquad \sin (x) &=&1 \\
&& \\
\sin ...
...
&& \\
x &=&\pi -\sin ^{-1}(1)\approx 1.570796 \\
\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...
...\
x_{3} &=&\sin ^{-1}\left( 1\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 6.2831853...
...
&& \\
x_{3} &\approx &1.5707963\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 $x=\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) \approx -0.52359878$

Left Side:

\begin{displaymath}2\sin^{2}\left( x\right) -\sin \left( x\right) -1\approx
2 \s...
...-0.52359878\right) -\sin \left( -0.52359878\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 -0.52359878 for x, then -0.52359878 is a solution.

Check solution $x=\pi -\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) \approx 3.665191$

Left Side:

\begin{displaymath}2\sin^{2}\left( x\right) -\sin \left( x\right) -1\approx
2 \s...
...left( 3.665191\right) -\sin \left( 3.665191\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 3.665191 for x, then 3.665191 is a solution.

Check solution $x=\sin ^{-1}\left( 1\right) \approx 1.5707963$

Left Side:

\begin{displaymath}2\sin^{2}\left( x\right) -\sin \left( x\right) -1\approx
2\si...
...ft( 1.5707963\right) -\sin \left( 1.5707963\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 1.5707963 for x, then 1.5707963 is a solution.

We have just verified that the exact solutions $x=\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) ,\pi -\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) $, and $\sin^{-1}\left( 1\right) $ are the solutions and these solutions repeat every $\pm 2\pi $ units. The approximate values of these solutions are$x\approx -0.52359878,\ 3.665191,$ and 1.5707963 and these solutions repeat every $\pm 6.2831853$ units.

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