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Mathematics LibreTexts

6.4 Solving Trig Equations

In Section 6.1, we determined the height of a rider on the London Eye Ferris wheel could be determined by the equation

If we wanted to know length of time during which the rider is more than 100 meters above ground, we would need to solve equations involving trig functions.

Solving using known values

In the last chapter, we learned sine and cosine values at commonly encountered angles. We can use these to solve sine and cosine equations involving these common angles.

 

Example 1

Solve for all possible values of t.

Notice this is asking us to identify all angles, t, that have a sine value of .  While evaluating a function always produces one result, solving for an input can yield multiple solutions.  Two solutions should immediately jump to mind from the last chapter:  and  because they are the common angles on the unit circle.

 

Looking at a graph confirms that there are more than these two solutions.  While eight are seen on this graph, there are an infinite number of solutions!

Remember that any coterminal angle will also have the same sine value, so any angle coterminal with these two is also a solution.  Coterminal angles can be found by adding full rotations of 2π, so we end up with a set of solutions:

 where k is an integer, and  where k is an integer

Example 2

A circle of radius  intersects the line x = -5 at two points.  Find the angles  on the interval , where the circle and line intersect.

 

The x coordinate of a point on a circle can be found as , so the x coordinate of points on this circle would be .  To find where the line x = -5 intersects the circle, we can solve for where the x value on the circle would be -5

                  Isolating the cosine

                                    Recall that , so we are solving

 

                    

 

We can recognize this as one of our special cosine values from our unit circle, and it corresponds with angles  and

Try it Now

  1. Solve  for all possible values of t.

 

Example 3

The depth of water at a dock rises and falls with the tide, following the equation , where t is measured in hours after midnight.  A boat requires a depth of 9 feet to tie up at the dock.   Between what times will the depth be 9 feet?

 

To find when the depth is 9 feet, we need to solve f(t) = 9.

               Isolating the sine

                     Dividing by 4

                       We know  when

 

While we know what angles have a sine value of , because of the horizontal stretch/compression, it is less clear how to proceed. 

To deal with this, we can make a substitution, defining a new temporary variable u to be , so our equation becomes

               

From earlier, we saw the solutions to this equation were

 where k is an integer, and

 where k is an integer

Undoing our substitution, we can replace the u in the solutions with  and solve for t

 

 where k is an integer, and   where k is an integer.

Dividing by π/12, we obtain solutions

 

 where k is an integer, and   where k is an integer.

 

The depth will be 9 feet and the boat will be able to approach the dock between 2am and 10am.

 

Notice how in both scenarios, the 24k shows how every 24 hours the cycle will be repeated.

In the previous example, looking back at the original simplified equation , we can use the ratio of the “normal period” to the stretch factor to find the period.  ;  notice that the sine function has a period of 24, which is reflected in the solutions: there were two unique solutions on one full cycle of the sine function, and additional solutions were found by adding multiples of a full period.

 

Try it Now

  1. Solve  for all possible values of t.

Solving using the inverse trig functions

The solutions to  do not involve any of the “special” values of the trig functions to we have learned.  To find the solutions, we need to use the inverse sine function.

Example 4

Use the inverse sine function to find one solution to .

 

Since this is not a known unit circle value, calculating the inverse, .  This requires a calculator and we must approximate a value for this angle.  If your calculator is in degree mode, your calculator will give you an angle in degrees as the output.  If your calculator is in radian mode, your calculator will give you an angle in radians.  In radians, , or in degrees, .

If you are working with a composed trig function and you are not solving for an angle, you will want to ensure that you are working in radians.  In calculus, we will almost always want to work with radians since they are unit-less.

Notice that the inverse trig functions do exactly what you would expect of any function – for each input they give exactly one output.  While this is necessary for these to be a function, it means that to find all the solutions to an equation like , we need to do more than just evaluate the inverse function.

 

Example 5
 
Example 5

Find all solutions to.

We would expect two unique angles on one cycle to have this sine value.  In the previous example, we found one solution to be .  To find the other, we need to answer the question “what other angle has the same sine value as an angle of 0.927?”  On a unit circle, we would recognize that the second angle would have the same reference angle and reside in the second quadrant.  This second angle would be located at , or approximately

To find more solutions we recall that angles coterminal with these two would have the same sine value, so we can add full cycles of 2π.

 

 and  where k is an integer,

or approximately,  and  where k is an integer.

Example 6

Find all solutions to  on the interval .

 

First we will turn our calculator to degree mode.  Using the inverse, we can find one solution .  While this angle satisfies the equation, it does not lie in the domain we are looking for.  To find the angles in the desired domain, we start looking for additional solutions. 

 

First, an angle coterminal with will have the same sine.  By adding a full rotation, we can find an angle in the desired domain with the same sine.

 

There is a second angle in the desired domain that lies in the third quadrant.  Notice that  is the reference angle for all solutions, so this second solution would be  past

 

The two solutions on  are x = and x =

Example 7

Find all solutions to  on .

 

Using the inverse tangent function, we can find one solution .  Unlike the sine and cosine, the tangent function only attains any output value once per cycle, so there is no second solution in any one cycle.

 

By adding π, a full period of tangent function, we can find a second angle with the same tangent value.  If additional solutions were desired, we could continue to add multiples of π, so all solutions would take on the form , however we are only interested in .

 

The two solutions on  are x = 1.249 and x = 4.391.

 

Try it Now

  1. Find all solutions to  on .

 

Example 8

Solve  for all solutions on one cycle,

 

         Isolating the cosine

              Using the inverse, we can find one solution

 

Thinking back to the circle, a second angle with the same cosine would be located in the third quadrant.  Notice that the location of this angle could be represented as .  To represent this as a positive angle we could find a coterminal angle by adding a full cycle.

 = 3.982

 

The equation has two solutions between 0 and 2π, at t = 2.301 and t = 3.982.

Example 9

Solve  for all solutions on two cycles, .

 

As before, with a horizontal compression it can be helpful to make a substitution, .  Making this substitution simplifies the equation to a form we have already solved.

 

A second solution on one cycle would be located in the fourth quadrant with the same reference angle.

In this case, we need all solutions on two cycles, so we need to find the solutions on the second cycle.  We can do this by adding a full rotation to the previous two solutions.

 

Undoing the substitution, we obtain our four solutions:

3t = 1.369, so t = 0.456

3t = 4.914 so t = 1.638

3t = 7.653, so t = 2.551

3t = 11.197, so t = 3.732

Example 10

Solve  for all solutions.

 

                                   Isolating the sine

                                    We make the substitution

                          Using the inverse, we find one solution

     

This angle is in the fourth quadrant.  A second angle with the same sine would be in the third quadrant with 0.730 as a reference angle:

 

We can write all solutions to the equation  as

 where k is an integer, or

 

Undoing our substitution, we can replace u in our solutions with  and solve for t

   or                             Divide by π

        or        

Try it Now

  1. Solve  for all solutions on one cycle, .

 

Solving Trig Equations

  1. Isolate the trig function on one side of the equation
  2. Make a substitution for the inside of the sine, cosine, or tangent (or other trig function)
  3. Use inverse trig functions to find one solution
  4. Use symmetries to find a second solution on one cycle (when a second exists)
  5. Find additional solutions if needed by adding full periods
  6. Undo the substitution

 

We now can return to the question we began the section with.

Example 11
 

The height of a rider on the London Eye Ferris wheel can be determined by the equation .  How long is the rider more than 100 meters above ground? 

 

To find how long the rider is above 100 meters, we first find the times at which the rider is at a height of 100 meters by solving h(t) = 100.

                       Isolating the cosine

                           We make the substitution

                                Using the inverse, we find one solution

 

         

This angle is in the second quadrant.  A second angle with the same cosine would be symmetric in the third quadrant.  This angle could be represented as u = -2.040, but we need a coterminal positive angle, so we add 2π:

 

Now we can undo the substitution to solve for t

so t = 9.740 minutes after the start of the ride

so t = 20.264 minutes after the start of the ride

 

A rider will be at 100 meters after 9.740 minutes, and again after 20.264.  From the behavior of the height graph, we know the rider will be above 100 meters between these times.  A rider will be above 100 meters for 20.265-9.740 = 10.523 minutes of the ride.

Important Topics of This Section

  • Solving trig equations using known values
  • Using substitution to solve equations
  • Finding answers in one cycle or period vs. finding all possible solutions
  • Method for solving trig equations

 

Try it Now Answers

  1.             
  2.  or
  3.  or

Section 6.4 Exercises

 

Find all solutions on the interval .

1.       2.          3.             4.

5.                6.                7.               8.

 

 

Find all solutions.

9.         10.         11.         12.

 

 

Find all solutions.

13.                      14.                  15.   

16.                   17.                     18.    

19.               20.                   21.      

22.                   23. .                   24.

 

Find all solutions on the interval .

25.         26.         27.       28.

29.      30.         31.         32.

 

Find the first two positive solutions.

33.         34.          35.       36.
37.       38.       39.       40.

Contributors

  • David Lippman (Pierce College)
  • Melonie Rasmussen (Pierce College)