Dawn with Advanced Functions..!!: 5.4 Solve Trigonometric Equation :

All about CALculations !!! :)

: http://www.youtube.com/watch?v=IE0FxGegdMg

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

Example 2:

$2\sin\left( 3x\right) -1=0$

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

$\begin{eqnarray*}&& \\ 2\sin \left( 3x\right) -1 &=&0 \\ && \\ \sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\ && \\ && \end{eqnarray*}$

If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2} \leq 3x\leq ,\displaystyle \frac{\pi }{2}\... ...[ -\displaystyle \frac{\pi }{6}\leq x\leq ,\displaystyle \frac{\pi }{6}\right]$ , we can use the inverse sine function to solve for reference angle 3x and then x.

$\begin{eqnarray*}&& \\ \sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\ &... ...&=&\sin ^{-1}\left( \displaystyle \frac{1}{2 }\right) \\ && \\ \end{eqnarray*}$

$\begin{eqnarray*}3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\ && ... ...\ x &\approx &0.174532925\ \mbox{ radians }\\ && \\ && \\ && \end{eqnarray*}$

We know that the $\sin$ e function is positive in the first and the second quadrant. Therefore two of the solutions are the angle 3x that terminates in the first quadrant and the angle $\pi -3x$ that terminates in the second quadrant. We have already solved for 3x.

$\begin{eqnarray*}&& \\ \sin \left( \pi -3x\right) &=&\displaystyle \frac{1}{2} ... ...-3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\ && \end{eqnarray*}$

$\begin{eqnarray*}&& \\ 3x &=&\pi -\sin ^{-1}\left( \displaystyle \frac{1}{2}\ri... ... \\ && \\ x &\approx &0.872665\ \mbox{ radians } \\ && \\ && \end{eqnarray*}$

The solutions are $x=\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right)$ and $x=\displaystyle \frac{\pi }{3}-\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) .\bigskip \bigskip\bigskip$
Example 3.
Solve the equation