Undoing our substitution, we can replace the u in the solutions with u =
t and solve
12
for t.
π π
π
π
t = + 2 k
π where k is an integer, and t 5
=
+ 2 k
π where k is an integer.
12
6
12
6
Dividing by π/12, we obtain solutions
t = 2+24 k where k is an integer, and
t = 10 + 24 k 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 24 k
shows how every 24 hours the cycle will
be repeated.
π
In the previous example, looking back at the original simplified equation
1
sin t = ,
12 2
we can use the ratio of the “normal period” to the stretch factor to find the period.
2π
12
= 2
π = 24 ; notice that the sine function has a period of 24, which is reflected
π
π
12
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
2. Solve 4sin(5 t)−1 = 1 for all possible values of t.
390 Chapter 6
Solving using the inverse trig functions
The solutions to sin(θ ) = 3
.
0 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 sin(θ ) = .
0 8.
Since this is not a known unit circle value, calculating the inverse, θ sin 1−
=
( 8.
0 ). 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, = sin 1−
θ
( 8.
0 ) ≈ 927
.
0
, or in degrees,
1
θ sin−
=
(0.8) ≈ 53.130° .
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 sin(θ ) = .
0 8, we need
to do more than just evaluate the inverse function.
Example 5
Find all solutions tosin(θ ) = 8
.
0 .
We would expect two unique angles on one cycle to have
0.8
this sine value. In the previous example, we found one
θ
solution to be = sin 1−
θ
( 8.
0 ) ≈ 927
.
0
. To find the other, we
0.929
need to answer the question “what other angle has the same
sine value as an angle of 0.927?” On a unit circle, we
1
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 θ = π sin 1−
−
( )
8
.
0 ,
or approximately θ ≈ π − 927
.
0
= 214
.
2
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.
θ =sin 1−(0 )8.+2 kπ and θ =π −sin 1−( )8.
0 + 2 k
π where k is an integer,
or approximately, θ = 927
.
0
+ 2 k
π and θ = 214
.
2
+ 2 k
π where k is an integer.
Section 6.4 Solving Trig Equations 391
Example 6
Find all solutions to
( x)
8
sin
= − on the interval °
0 ≤ x <
°
360 .
9
First we will turn our calculator to degree mode. Using the inverse, we can find one
solution
8
=
−
sin 1
x
− ≈ −
734
.
62
° . While this angle satisfies the equation, it does not
9
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 −
°
734
.
62
will have the same sine. By adding a full
rotation, we can find an angle in the desired domain with the same sine.
x = −
°
734
.
62
+
°
360 =
°
266
.
297
There is a second angle in the desired domain that lies in the third quadrant. Notice that
.
62
°
734 is the reference angle for all solutions, so this second solution would be
.
62
°
734 past
°
180
x = .
62
°
734 +
°
180 =
°
734
.
242
The two solutions on 0°≤ x < °
360 are x =
°
266
.
297
and x =
°
734
.
242
Example 7
Find all solutions to tan( x) = 3 on 0 ≤ x < π
2 .
Using the inverse tangent function, we can find one solution = tan 1−
x
(3) ≈1 249
.
.
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
1.249
solutions were desired, we could continue to add multiples
of π, so all solutions would take on the form
1
x = .
1 249 + π
k , however we are only interested in
4.391
0 ≤ x < π
2 .
x = .
1 249 + π = .
4 391
The two solutions on 0≤ x < π2 are x = 1.249 and x = 4.391.
Try it Now
3. Find all solutions to tan ( x) = 0.7 on °
0 ≤ x <
°
360 .
392 Chapter 6
Example 8
Solve 3cos( t)+ 4 = 2 for all solutions on one cycle, 0 ≤ t < 2π
3cos( t)+4= 2 Isolating the cosine
3cos( t) = 2
−
( t)
2
cos = −
Using the inverse, we can find one solution
3
−
2
t = cos 1
− ≈ .
2 301
3
2.301
Thinking back to the circle, a second angle with the same
2
−
cosine would be located in the third quadrant. Notice that
3
the location of this angle could be represented as
1
-2.301
t =
301
.
2
−
. To represent this as a positive angle we could
or 3.982
find a coterminal angle by adding a full cycle.
t = − 301
.
2
+ π
2 = 3.982
The equation has two solutions between 0 and 2π, at t = 2.301 and t = 3.982.
Example 9
π
Solve cos(3 t) = 2
.
0 for all solutions on two cycles,
4
0 ≤ t <
.
3
As before, with a horizontal compression it can be helpful to make a substitution,
u = t
3 . Making this substitution simplifies the equation to a form we have already
solved.
cos( u) = .
0 2
= cos 1−
u
( 2.
0 ) ≈ 369
.
1
A second solution on one cycle would be located in the fourth quadrant with the same
reference angle.
u = 2π −1 369
.
= 914
.
4
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.
u = 369
.
1
+ 2π = 653
.
7
u = 4 914
.
+ 2π =
197
.
11
Undoing the substitution, we obtain our four solutions:
3 t = 1.369, so t = 0.456
3 t = 4.914 so t = 1.638
3 t = 7.653, so t = 2.551
3 t = 11.197, so t = 3.732
Section 6.4 Solving Trig Equations 393
Example 10
Solve 3sin(π t) = 2
− for all solutions.
3sin(π t)= 2−
Isolating the sine
(π t)
2
sin
= −
We make the substitution u = π t
3
( u)
2
sin
= −
Using the inverse, we find one solution
3
−
2
u = sin 1
− ≈ − .
0 730
3
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:
u = π + 0 730
.
= 871
.
3
We can write all solutions to the equation
( u)
2
sin
= − as
3
u = 0
− 730
.
+ π
2 k where k is an integer, or
u = 871
.
3
+ π
2 k
Undoing our substitution, we can replace u in our solutions with u =π t and solve for t π t = 0
− 730
.
+ π
2 k or
π t = .
3 871+ π
2 k
Divide by π
t = − 232
.
0
+ 2 k
or
t =1.232 + 2 k
Try it Now π
4. Solve 5sin
t + 3 = 0 for all solutions on one cycle, 0 ≤ t < π
2 .
2
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.
394 Chapter 6
Example 11
The height of a rider on the London Eye Ferris wheel can be determined by the equation
π
h( t) = − .
67 5cos
t + .
69 5 . How long is the rider more than 100 meters above
15
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.
π
100 = − 5
.
67 cos
t + .
69 5
Isolating the cosine
15
π
5
.
30 = −67 5
. co
s
t
15
5
.
30
π
π
= co
s
t
We make the substitution u =
t
− .
67 5
15
15
5
.
30
= cos( u)
Using the inverse, we find one solution
−
5
.
67
−
.
30 5
u = cos 1
≈ 040
.
2
− 67 5
.
This angle is in the second quadrant. A second angle with
u = 2.040
the same cosine would be symmetric in the third quadrant.
This angle could be represented as u = -2.040, but we need a
1
coterminal positive angle, so we add 2π:
u = -2.040
u = 2π − .
2 040 ≈ 244
.
4
or 4.244
Now we can undo the substitution to solve for t
π t = 040
.
2
so t = 9.740 minutes after the start of the ride
15
π t = 244
.
4
so t = 20.264 minutes after the start of the ride
15
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
Section 6.4 Solving Trig Equations 395
Try it Now Answers
π
1.
+π k
4 π π
π
π
2. t
2
=
+
k t
2
=
+
k
30
5
6
5
3. x =
°
992
.
34
or x =
°
180 + .
34 °
99 =
.
214
°
992
4. t = 3.590 or t = 2.410
396 Chapter 6
Section 6.4 Exercises
Find all solutions on the interval 0≤θ <2π .
1. 2sin (θ ) = − 2 2. 2sin (θ ) = 3
3. 2cos(θ ) =1
4. 2cos(θ ) = − 2
5. sin (θ ) =1
6. sin (θ ) = 0
7. cos(θ ) = 0
8. cos(θ ) = 1
−
Find all solutions.
9. 2cos(θ ) = 2
10. 2cos(θ ) = 1
−
11. 2sin (θ ) = 1
−
12. 2sin (θ ) = − 3
Find all solutions.
13. 2sin (3θ ) =1
14. 2sin (2θ ) = 3
15. 2sin (3θ ) = − 2
16. 2sin (3θ ) = 1
−
17. 2cos(2θ ) =1
18. 2cos(2θ ) = 3
π
19. 2cos(3θ ) = − 2
20. 2cos(2θ ) = 1
−
21. cos
θ =
1
−
4
π
π
22. sin
θ =
1
−
23. 2sin (πθ ) =1.
24. 2cos
θ =
3
3
5
Find all solutions on the interval 0≤ x<2π.
25. sin ( x) = 0.27
26. sin ( x) = 0.48
27. sin ( x) = 0.58
−
28. sin ( x) = 0.34
−
29. cos( x) = 0.55
−
30. sin ( x) = 0.28
31. cos( x) = 0.71
32. cos( x) = 0.07
−
Find the first two positive solutions.
33. 7sin (6 x) = 2
34. 7sin (5 x) = 6
35. 5cos(3 x) = 3
− 36. 3cos(4 x) = 2
π
π
π
π
37. 3sin
x =
2 38. 7sin
x =
6 39. 5cos
x =
1 40. 3cos
x = 2
−
4
5
3
2
Section 6.5 Modeling with Trigonometric Functions 397
Section 6.5 Modeling with Trigonometric Functions
Solving right triangles for angles
In Section 5.5, we used trigonometry on a right triangle to solve for the sides of a triangle
given one side and an additional angle. Using the inverse trig functions, we can solve for
the angles of a right triangle given two sides.
Example 1
An airplane needs to fly to an airfield located 300 miles east and 200 miles north of its
current location. At what heading should the airplane fly? In other words, if we ignore
air resistance or wind speed, how many degrees north of east should the airplane fly?