SOLVING TRIGONOMETRIC EQUATIONS

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

Example 4:        Solve for x in the following equation.

$$\sin ^{2}\left( 2x\right) -\sin \left( 2x\right) +\displaystyle \displaystyle \frac{1}{4}=0$$

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

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

If we restrict the domain of the sine function to $\left[ -\displaystyle \displaystyle \frac{\pi }{2}% \leq 2x\leq \displaystyle \... ...yle \frac{\pi }{4}\leq x\leq \displaystyle \displaystyle \frac{\pi }{4}\right]$, we can use the inverse sine function to solve for reference angle x, and then x. The reference angle is always in the first quadrant and positive.

$$\begin{array}{rclll} && \\ \sin \left( 2x\right) &=&\display... ...}{12} \\ && \\ x_{1} &\approx &0.261799387799 \\ \end{array}$$

The $\sin \left( 2x\right)$ has a period of $\pi$. This means that it makes one rotation every $\pi$ radians. We can divide the interval $\left[ 0,\pi \right]$ into four parts: quadrant I, quadrant II, quadrant III, and quadrant IV. The sine function is positive in the first and second quadrants and negative in the third and and fourth quadrants. In this case, one-fourth of the rotation is We will use the reference angle $x^{\prime }$ to find the four angles.

The first solution is the angle that terminates in the first quadrant: is $% x_{1}=x^{\prime }=\displaystyle \displaystyle \frac{\pi }{12}.$ The second solution is the angle that terminates in the second quadrant: $x_{2}=\displaystyle \displaystyle \frac{\pi }{2}-x^{\prime }=\displaystyle \dis... ... \frac{\pi }{12}=\displaystyle \displaystyle \frac{5\pi }{12}.\bigskip\bigskip$

The period of the $\sin \left( 2x\right)$ function is $\pi .$ This means that the values will repeat every $\pi$ radians in both directions. Therefore, the exact solutions are

$$\begin{array}{rclll} && \\ x &=&\displaystyle \displaystyle ... ...c{5\pi }{12}\pm n\left( \pi \right) , \\ && \\ && \end{array}$$

where n is an integer.

The approximate solutions are $x\approx 0.261799387799\pm n\left( \pi \right)$ and $x\approx 1.308996939\pm n\left( \pi \right)$ 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=0.261799387799

Left Side:

$$\begin{array}{rclll} \sin ^{2}\left( 2x\right) -\sin \left( 2... ...le \displaystyle \frac{1}{4} \\ && \\ &\approx &0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute <tex2html_comment_mark> 0.261799387799 for x, then 0.261799387799 is a solution.

Check answer . x=1.308996939

Left Side:

$$\begin{array}{rclll} \sin ^{2}\left( 2x\right) -\sin \left( 2... ...le \displaystyle \frac{1}{4} \\ && \\ &\approx &0 \end{array}$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 1.308996939for x, then 1.308996939 is a solution.

Graphical Check:

Graph the equation $f\left( x\right) =\sin ^{2}\left( 2x\right) -\sin \left( 2x\right) +\displaystyle \displaystyle \frac{1}{4}.$ Note that the graph crosses the x-axis many times indicating many solutions.

Note that it crosses two times in the interval from 0 ro $\pi :0.261799387799$ and 1.308996939.

Since the period is $\pi$, the graph crosses again at $0.261799387799\pm \pi$ and $1.308996939\pm \pi$ etc.

If you would like to test yourself by working some problems similar to this example, click on Problem.

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