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Solve for x in the following equation.

Problem 9.2a:

$8\sin\left( 4x\right) -3=0$

Answers:         There are an infinite number of solutions:

$x=\displaystyle \frac{1}{4}
\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \pm n\left( \displaystyle \frac{\pi }{2}\right) $and $x=\displaystyle \frac{\pi }{4}-\displaystyle \frac{1}{4}\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \pm
\pi \left( \displaystyle \frac{\pi }{2}\right) $ are the exact solutions, and $x\approx
0.096099\pm n\left( \displaystyle \frac{\pi }{2}\right) $ and $x\approx 0.689299\pm
n\left( \displaystyle \frac{\pi }{2}\right) $ are the approximate solutions.


To solve for x, first isolate the sine term.

\begin{eqnarray*}&& \\
8\sin\left( 4x\right) -3 &=&0 \\
&& \\
\sin\left( 4x\right) &=&\displaystyle \frac{3}{8} \\
&& \\
&& \\

If we restrict the domain of the cosine function to $-\displaystyle \frac{\pi }{2}\leq
4x\leq \displaystyle \frac{\pi }{2}\rightarrow -\displaystyle \frac{\pi }{8}\leq x\leq \displaystyle \frac{\pi }{8}$, we can use the arcsin function to solve for x.

\begin{eqnarray*}\sin\left( 4x\right) &=&\displaystyle \frac{3}{8} \\
&& \\
\s... &=&\sin ^{-1}\left( \displaystyle \frac{3}{8}
\right) \\
4x &=&\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \...{3}{8}\right) \\
&& \\
x &\approx &0.096099193 \\
&& \\

The sine of x is positive in the first quadrant and the second quadrant. This means that there are two solutions in the first counterclockwise rotation from 0 to $2\pi $. One angle 4x terminates in the first quadrant and the second angle terminates in the second quadrant. One solution is $x=\displaystyle \frac{1}{4}\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) $

The period of $\sin x$ is $2\pi $, and the period of $\sin 4x$ is $\displaystyle \frac{
\pi }{2}.$ Therefore,the second solution is $x=\displaystyle \frac{\pi }{4}-\displaystyle \frac{1}{4}
\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \approx 0.689299.\bigskip\bigskip $

Since the period is $\displaystyle \frac{\pi }{2},$ this means that the values will repeat every $\displaystyle \frac{\pi }{2}$ radians. Therefore, the solutions are $
x\approx 0.096099193\pm n\left( \displaystyle \frac{\pi }{2}\right) $ and $x=\pm n\left(
\displaystyle \frac{\pi }{2}\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 the answer x=0.096099193

Since the left side equals the right side when you substitute 0.096099193for x, then 0.096099193 is a solution.

Check the answer x=0.689299

Since the left side equals the right side when you substitute 0.689299 for x, then 0.689299 is a solution.

Graphical Check:

Graph the equation

$f(x)=8\sin\left( 4x\right) -3.$

Note that the graph crosses the x-axis many times indicating many solutions.

Note the graph crosses at 0.096099193 ( one of the solutions ). Since the period of the function is $\displaystyle \frac{\pi }{2}\approx
1.5707963$, the graph crosses again at 0.096099193+1.5707963=1.6668955 and again at $0.096099193+2\left( 1.5707963\right) \approx 3.237691793$, etc.

The graph also crosses at 0.689299 ( another solution we found ). Since the period is $\displaystyle \frac{\pi }{2}\approx
1.5707963$, it will crosses again at 0.689299+1.5707963=2.2600953 and at $0.689299+2\left(
1.5707963\right) =3.83089$, etc $.\bigskip\bigskip $

If you would like to review the solution to problem 9.2b, click on solution.

If you would like to go back to the previous section, click on previous

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