SOLVING TRIGONOMETRIC EQUATIONS

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

Example 1:        Solve for x in the following equation.

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

There are an infinite number of solutions to this problem.

Isolate the sine term. To do this, rewrite the left side of the equation in an equivalent factored form.

$$\begin{array}{rclll} 2\sin^{2}\left( x\right) -\sin \left( x\... ...\sin (x)+1\right) \left( \sin (x)-1\right) &=&0 \\ \end{array}$$

The product of two factors equals zero if at least one of the factors equals zeros. This means that $\left( 2\sin (x)+1\right) \left( \sin (x)-1\right) =0\$if $2\sin (x)+1=0$ or $\sin (x)-1=0.$

We just transformed a difficult problem into two easier problems. To find the solutions to the original equation, $\left( 2\sin (x)+1\right) \left( \sin (x)-1\right) =0\$, we find the solutions to the equations $2\sin (x)+1=0$and $\sin (x)-1=0.$

$$\begin{array}{rclll} 2\sin (x)+1 &=&0 \\ && \\ \sin \left( x\right) &=&-\displaystyle \frac{1}{2} \\ \end{array}$$

and

$$\begin{array}{rclll} \sin (x)-1 &=&0 \\ && \\ \sin (x) &=&1 \\ \end{array}$$

How do we isolate the x? We could take the arcsine of both sides. However, the sine function is not a one-to-one function.

Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. The graph of the sine function is one-to-one on the interval $\left[ -\displaystyle \frac{\pi }{2}, \displaystyle \frac{\pi }{2}\right] .$ If we restrict the domain of the sine function to that interval , we can take the arcsine of both sides of each equation.

$$\begin{array}{rclll} (1)\qquad \sin \left( x\right) &=&-\disp... ...aystyle \frac{1}{2}\right) \approx -0.523598776 \\ \end{array}$$

We know that $\sin \left( x\right) =\sin \left( \pi -x\right) .$ Therefore, if $\sin (x)=-\displaystyle \frac{1}{2}$, then $\sin (\pi -x)=-\displaystyle \frac{1}{2}.$

$$\begin{array}{rclll} \left( 2\right) \qquad \sin (\pi -x) &=&... ...isplaystyle \frac{1}{2}\right) \approx 3.665191 \\ \end{array}$$

We complete the problem by solving for the second factor.

$$\begin{array}{rclll} (3)\qquad \sin (x) &=&1 \\ && \\ \sin ... ... && \\ x &=&\pi -\sin ^{-1}(1)\approx 1.570796 \\ \end{array}$$

Since the period of $\sin (x)$ equals $2\pi$, these solutions will repeat every $2\pi$ units. The exact solutions are

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

where n is an integer.

The approximate values of these solutions are

$$\begin{array}{rclll} x_{1} &\approx &-0.52359878\pm 6.2831853... ... && \\ x_{3} &\approx &1.5707963\pm 6.2831853n \\ \end{array}$$

where n is an integer.

One can check each solution algebraically by substituting each solution in the original equation. If, after the substitution, the left side of the original equation equals the right side of the original equation, the solution is valid.

One can also check the solutions graphically by graphing the function formed by the left side of the original equation and graphing the function formed by the right side of the original equation. The x-coordinates of the points of intersection are the solutions. The right side of the equation is 0 and f(x)=0 is the x-axis. So really what you are looking for are the x-intercepts to the function formed by the left side of the equation.

Algebraic Check:

Check solution $x=\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) \approx -0.52359878$

Left Side:

$$2\sin^{2}\left( x\right) -\sin \left( x\right) -1\approx 2 \s... ...-0.52359878\right) -\sin \left( -0.52359878\right) -1\approx 0$$

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute -0.52359878 for x, then -0.52359878 is a solution.

Check solution $x=\pi -\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) \approx 3.665191$

Left Side:

$$2\sin^{2}\left( x\right) -\sin \left( x\right) -1\approx 2 \s... ...left( 3.665191\right) -\sin \left( 3.665191\right) -1\approx 0$$

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute 3.665191 for x, then 3.665191 is a solution.

Check solution $x=\sin ^{-1}\left( 1\right) \approx 1.5707963$

Left Side:

$$2\sin^{2}\left( x\right) -\sin \left( x\right) -1\approx 2\si... ...ft( 1.5707963\right) -\sin \left( 1.5707963\right) -1\approx 0$$

Right Side:        0

Since the left side of the original equation equals the right side of the original equation when you substitute 1.5707963 for x, then 1.5707963 is a solution.

We have just verified that the exact solutions $x=\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right) ,\pi -\sin ^{-1}\left( -\displaystyle \frac{1}{2}\right)$, and $\sin^{-1}\left( 1\right)$ are the solutions and these solutions repeat every $\pm 2\pi$ units. The approximate values of these solutions are $x\approx -0.52359878,\ 3.665191,$ and 1.5707963 and these solutions repeat every $\pm 6.2831853$ units.

Graphical Check:

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

The graph crosses the x-axis at -0.52359878. Since the period is $2\pi \approx 6.2831853$, you can verify that the graph also crosses the x-axis again at -0.52359878+6.2831853=5.7595865 and at $-0.52359878+2\left(6.2831853\right) =12.04277$, etc.

The graph crosses the x-axis at $\ 3.665191$. Since the period is $2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2html_comment_mark> 3.665191+6.2831853=9.9483763 and at $3.665191+2\left( 6.2831853\right) =16.2315616$, etc..

The graph crosses the x-axis at $\ 1.5707963$. Since the period is $2\pi \approx 6.2831853$, the graph also crosses the x-axis again at <tex2html_comment_mark> 1.5707963+6.2831853=7.8539816 and at $1.5707963+2\left( 6.2831853\right) =14.13716694$, etc.

Note: If the problem were to find the solutions in the interval $\left[ 0,2\pi \right]$, then you choose those solutions from the set of infinite solutions that belong to the set $\left[ 0,2\pi \right] :$ $x\approx 1.5707963,\quad 3.665191$, and 5.7595865.

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.

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

Copyright © 1999-2024 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour