SOLVING TRIGONOMETRIC EQUATIONS

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

Problem 9.4a: Solve for x in the following equation.

$\begin{displaymath}3\sin ^{2}x-2=0\end{displaymath}$

Answers: There are an infinite number of solutions: $x=\sin ^{-1}\sqrt{% \displaystyle \displaystyle \frac{2}{3}}\pm n\left( 2\pi \right) ,$ $x=\pi -\sin ^{-1}\sqrt{\displaystyle \displaystyle \frac{2}{3}% }\pm n\left( 2\pi \right) ,$ $x=\pi +\sin ^{-1}\sqrt{\displaystyle \displaystyle \frac{2}{3}}\pm n\left( 2\pi \right) ,$ and $x=2\pi -\sin ^{-1}\sqrt{\displaystyle \displaystyle \frac{2}{3}}\pm n\left( 2\pi \right)$ are the exact solutions, and $x\approx 0.9553166\pm n\left( 2\pi \right) ,$ $x\approx 2.186276\pm n\left( 2\pi \right) ,$ $% x\approx 4.096909\pm n\left( 2\pi \right) ,$ and $x\approx 5.327869\pm n\left( 2\pi \right)$ are the approximate solutions.

Solution:

To solve for x, first isolate the sine term.

$\begin{displaymath}\begin{array}{rclll} && \\ 3\sin ^{2}x-2 &=&0 \\ && \\ \si... ...ght) &=&\pm \sqrt{\displaystyle \frac{2}{3}} \\ && \end{array}\end{displaymath}$

If we restrict the domain of the sine function to $-\displaystyle \displaystyle \frac{\pi }{2}\leq x\leq \displaystyle \displaystyle \frac{\pi }{2}$ , we can use the arcsin function to solve for the reference angle $x^{\prime }$ , and then x. The reference angle is always in the first quadrant.

$\begin{displaymath}\begin{array}{rclll} && \\ \sin \left( x\right) &=&\sqrt{\di... ...{2}{3}}\right) \\ && \\ x &\approx &0.9553166 \\ \end{array}\end{displaymath}$

The sine of x is positive in the first and second quadrant and negative in the third quadrant and the fourth quadrant. This means that there are four solutions in the first counterclockwise rotation from 0 to $2\pi$ .

One angle, $x_{1}=x^{\prime }=\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right) \approx 0.9553166$ terminates in the first quadrant, a second angle $% x_{2}=\pi -x^{\prime }=\pi -\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right) \approx 2.186276$ terminates in the second quadrant, a third angle $% x_{3}=\pi +x^{\prime }=\pi +\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right) \approx 4.096909$ terminates in the third quadrant, and a fourth angle $% x_{4}=2\pi -x^{\prime }=2\pi -\sin ^{-1}\left( \sqrt{\displaystyle \displaystyle \frac{2}{3}}\right) \approx 5.327869$ terminates in the fourth quadrant.

Since the period is $2\pi ,$ this means that the values will repeat every $% 2\pi$ radians. Therefore, the solutions are

$\begin{displaymath}\begin{array}{rclll} && \\ x &=&\sin ^{-1}\left( \sqrt{\disp... ...}{3}}\right) \pm n\left( 2\pi \right) \\ && \\ && \end{array}\end{displaymath}$

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=0.9553166

Left Side:

$\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 0.9553166\right) -2\approx 0\bigskip\end{displaymath}$

Right Side: $0\bigskip$

Since the left side equals the right side when you substitute 0.9553166for x, then 0.9553166 is a solution.

Check the answer . x=2.186276

Left Side:

$\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 2.186276\right) -2\approx 0\bigskip\end{displaymath}$

Right Side: $0\bigskip$

Since the left side equals the right side when you substitute 2.186276 for x, then 2.186276 is a solution.

Check the answer . x=4.096909

Left Side:

$\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 4.096909\right) -2\approx 0\bigskip\end{displaymath}$

Right Side: $0\bigskip$

Since the left side equals the right side when you substitute 4.096909 for x, then 4.096909 is a solution.

Check the answer . x=5.327869

Left Side:

$\begin{displaymath}3\sin ^{2}x-2=3\sin ^{2}\left( 5.327869\right) -2\approx 0\bigskip\end{displaymath}$

Right Side: $0\bigskip$

Since the left side equals the right side when you substitute 5.327869 for x, then 5.327869 is a solution.

Graphical Check:

Graph the equation $f(x)=3\sin ^{2}x-2.$ (Formed by subtracting the right side of the original equation from the left side of the original equation.

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

Note the graph crosses at 0.9553166, 2.186276, 4.096909, and 5.327869 . Since the period is $2\pi \approx 6.283185$ , the graph will cross the x-axis every $2\pi$ units to the left and right of each number.

If you would like to review problem 9.4b, click on problem 9.4b.

If you would like to go to the next section, click on Next.

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

If you would like to go back to the equation table of contents, click on Contents.

[Algebra] [Trigonometry] [Geometry] [Differential Equations] [Calculus] [Complex Variables] [Matrix Algebra]
S.O.S MATHematics home page

Author: Nancy Marcus