SOLVING TRIGONOMETRIC EQUATIONS

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

Example 2:        Solve for x in the following equation.

$$4\cos ^{2}x-3=0$$

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

$$\begin{array}{rclll} && \\ 4\cos ^{2}x-3 &=&0 \\ && \\ 4\c... ...tyle \displaystyle \frac{\sqrt{3}}{2} \\ && \\ && \end{array}$$

If we restrict the domain of the cosine function to $\left[ 0\leq x\leq \pi \right]$, we can use the inverse cosine function to solve for reference angle $x^{\prime }$, and then x. The reference angle is always located in the first quadrant and positive.

$$\begin{array}{rclll} && \\ \cos x &=&\displaystyle \displays... ...{2}\right) \\ && \\ x &\approx &0.52359877 \\ && \end{array}$$

The cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants. 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 $% x_{1}=x^{\prime }= \cos ^{-1}\left( \displaystyle \displaystyle \frac{\sqrt{3}}{2}\right) .$ The second solution is the angle that terminates in the second quadrant: $x_{2}=\pi -x^{\prime }=\pi - \cos ^{-1}\left( \displaystyle \displaystyle \frac{\sqrt{3}}{2}\right)$. The third solution is the angle that terminates in the third quadrant: $x_{3}=\pi + \cos ^{-1}\left( \displaystyle \displaystyle \frac{\sqrt{3}}{2}\right) .$ The fourth solutions is the angle that terminates in the fourth quadrant: $2\pi - \cos ^{-1}\left( \displaystyle \displaystyle \frac{% \sqrt{3}}{2}\right) .\bigskip\bigskip$

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

$$\begin{array}{rclll} && \\ x &=& \cos ^{-1}\left( \displayst... ...qrt{3}}{2}\right) \pm n\left( 2\pi \right) , \\ && \end{array}$$

where n is an integer.

The approximate solutions are $x\approx 0.52359877\pm n\left( 2\pi \right) ,$ $x\approx 2.6179938\pm n\left( 2\pi \right)$, $x\approx 3.66519\pm n\left( 2\pi \right)$ and $x\approx 5.759586537\pm n\left( 2\pi \right)$ 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=0.52359877

Left Side:

$$4\cos ^{2}x-3\approx 4\cos ^{2}\left( 0.52359877\right) -3\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 0.52359877for x, then 0.52359877 is a solution.

Check answer . x=2.6179938

Left Side:

$$4\cos ^{2}x-3\approx 4\cos ^{2}\left( 2.6179938\right) -3\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 2.6179938for x, then 2.6179938 is a solution.

Check answer .x=3.66519

Left Side:

$$4\cos ^{2}x-3\approx 4\cos ^{2}\left( 3.66519\right) -3\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 3.66519 for x, then 3.66519 is a solution.

Check answer . x=5.759586537

Left Side:

$$4\cos ^{2}x-3\approx 4\cos ^{2}\left( 5.759586537\right) -3\approx 0\bigskip$$

Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 5.759586537for x, then 5.759586537 is a solution.

Graphical Check:

Graph the equation $4\cos ^{2}x-3.$ Note that the graph crosses the x-axis many times indicating many solutions.

Note that it crosses four time in the interval from 0 ro $2\pi :0.52359877,\ 2.6179938$, 3.66519 and 5.759586537.

Since the period is $2\pi$, the graph crosses again at $0.52359877+2\pi ,\ 2.617899$ $+2\pi$, $3.66519+2\pi$, and $5.759586537+2\pi$ etc.

If you would like to work another example, click on Example.

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

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