Solving Trig Equations: The Easy Ones - MathHands
Trigonometry Sec. 01 notes
MathHands.com M´ arquez
Solving Trig Equations: The Easy Ones Main Idea We are now ready to discuss the solving of trigonometric equations. Recall that, generally speaking, identities are equations which hold true for all values. Today, we consider equations that, generally speaking, are only true for some values of x (or θ, or any variable/s). Our goal is to determine exactly for which values the equation holds true. This is the general meaning of solving equations. For example, consider the equation, x2 = 1 Clearly, this equation is not true for all real values of x. Nevertheless, we can solve it by taking careful steps:
x2 = 1
(given)
x −1=0 (x − 1)(x + 1) = 0
(algebra) (factor, algebra)
2
x−1=0 or x+1=0 x=1 or x = −1
(Zero Factor Theorem) (algebra)
Thus, this equation has two real solutions. Let us call the two solutions x1 and x2 . Then the solutions are x1 = 1 and x2 = −1
This is a classic polynomial equation, hopefully one you find familiar from previous studies. Here, we solve equations that involve the variables on the trigonometric functions, thus the description trig equations. We have actually already taken a first look at such equations. Recall we have seen equations such as sin(x) =
3 4
when we first introduced the inverse trig function, sin−1 r, or arcsin r. This was particularly useful in solving right triangles when a ratio was known, such as 3/4, and the angle was sought. At that point, we used a calculator to compute/estimate the arcsine of 3/4, in degrees mode, as sin−1 43 ≈ 48.6◦ . At that point we also talked about the domain restriction and the fact that the domain for sin−1 is only real rations, r where −1 ≤ r ≤ 1 while the codomain is only angles θ where −90◦ ≤ θ90◦ . We claimed we would come back and revisit such equations to solve completely. Well.. the time is now and the place is here. We now have all the necessary time and tools to solve, completely, correctly, and honestly, trig equations such as sin(x) =
3 4
The general strategy we will employ is to graph both sides of the equation. In other words, we will graph y = sin x. We will then graph on the same plane, y = 34 . We will then see if and where these two graphs intersect. They will intersect exactly where sin(x) = 34 . At that point, we simply have to point to the solutions and write them down. This last step may take some getting used to because, unlike x2 = 1, trig equations may have infinite many solutions.... talk is cheap.. let us solve!
c
2007-2009 MathHands.com
math hands
pg. 1
Trigonometry Sec. 01 notes
MathHands.com M´ arquez
Example Solve sin (x) =
3 4
Solution: First we graph y = 3/4 then we graph y = sin x, then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them.
2.5 2.0 y=
3 4
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.59
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0 -1.5 -2.0 -2.5
Now, we need to describe them. To do so, it may be sometimes helpful to separate all the solutions into two different ’buckets’. On one bucket we place all the ’uphill solutions’, namely the ones that occur as the sine graph increases.
c
2007-2009 MathHands.com
math hands
pg. 2
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y=
3 4
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.59
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0 -1.5 ≈ −671.41◦
360◦
≈ −311.2◦
360◦ -2.0
≈ 48◦
360◦
≈ 408◦
360◦
-2.5
Of these, the first one is determined, when the ratio is not a famous one such as 3/4, by using a calculator to estimate sin−1 (3/4) ≈ 48.59◦. This solution is the first one that occurs, right of the origin. we could call this, our fist solution, x1 , the next one occurs exactly 360◦ right of that one, x2 ≈ 408.59◦ and so on.. While on the left side, we find a solution at x−1 ≈ 48.59◦ − 350◦ = −311.41. In other words, the uphill solutions are given by a clear pattern: . . . , −311.41◦, 48.59◦, 408.59◦, . . . In fact, a more descriptive and concise way to describe such sequence is to use a closed formula that describes the pattern. Namely, we can describe all such uphill solutions by a formula. The formula would have to say that we can start with 48.59◦ (or any other uphill solution) then proceed to add or subtract any whole number multiple of 360◦ . If we allow k to be any integer [we write k ∈ Z], positive or negative, this can be accomplished by the following formula: xk ≈ 48.59◦ + k360◦
for k ∈ Z
Indeed, this is a common and effective way to describe the uphill solutions. Note one could also use radians, xk ≈ 0.848 + k2π
for k ∈ Z
This sets up a correspondence, for every integer value k, xk is an uphill solution, and every uphill solution corresponds to exactly one value of k. It should be noted that sometimes, authors leave out the k, and it is understood that each value of k will produce a corresponding value solution x, thus the following notation is common: x ≈ 48.59◦ + k360◦
for k ∈ Z
Of course, this is only half of the solutions, now we turn our attention to the downhill solutions, and proceed in a similar manner.
c
2007-2009 MathHands.com
math hands
pg. 3
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y=
3 4
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.59
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0 -1.5 ≈ −671.41◦
≈ −311.2◦
360◦
360◦ -2.0
≈ −588
◦
360
◦
≈ −228
◦
≈ 48◦
360◦ -2.5
360◦
≈ 132◦
≈ 408◦ 360◦
≈ 492◦
360◦ 360◦
We use the symmetry of the graph to determine one of the solutions. For example, the first solution right of the origin is ≈ 180◦ − 48.59◦, and in this case as well, all the downhill solutions are equally spaced 360◦ apart, thus we can write a closed form for all the downhill solutions: xk ≈ 131.41◦ + k360◦
for k ∈ Z
We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 48.59◦ + k360◦
for k ∈ Z
xk ≈ 131.41◦ + k360◦
for k ∈ Z
OR
Example Solve sin (x) =
1 3
Solution: First we graph y = 0.333 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (0.333) ≈ 19.471◦. This solution is the only one provided by the sin−1 function.
c
2007-2009 MathHands.com
math hands
pg. 4
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y=
1 3
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.471
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0
≈ −700.529◦
360◦
≈ −340.30◦
360◦
-1.5 ≈ 19.28◦
360◦
-2.0 ≈ −559
◦
360
◦
≈ −199
◦
360◦ -2.5
≈ 161◦
≈ 380.8◦ 360◦
360◦ ≈ 521◦
360◦
We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 19.471◦ + k360◦
for k ∈ Z
xk ≈ 160.529◦ + k360◦
for k ∈ Z
OR
Example Solve sin (x) = −
3 5
Solution: First we graph y = −0.6 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (−0.6) ≈ −36.87◦. This solution is the only one provided by the sin−1 function.
c
2007-2009 MathHands.com
math hands
pg. 5
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y= −
3 5
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.13
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0
≈ −756.87◦
360◦
≈ −396.8◦
-1.5 ≈ −36◦
360◦
360◦
-2.0 ≈ −503
◦
360
◦
≈ −143
360◦
◦
-2.5
≈ 217◦
≈ 324◦ 360◦
360◦ ≈ 577◦
360◦
We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ −36.87◦ + k360◦
for k ∈ Z
xk ≈ 216.87◦ + k360◦
for k ∈ Z
OR
c
2007-2009 MathHands.com
math hands
pg. 6
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solving Trig Equations: The Easy Ones
1. Solve sin (x) =
2 5
sin (x) =
4 5
sin (x) =
6 7
2. Solve
3. Solve
4. Solve
5 11
sin (x) = − 5. Solve cos (x) =
7 10
6. Solve
7 10
cos (x) = − 7. Solve cos (x) =
3 5
cos (x) =
9 10
8. Solve
9. Solve cos (x) = −
8 15
cos (x) = −
3 4
tan (x) = −
3 4
10. Solve
11. Solve
12. Solve tan (x) =
3 5
tan (x) =
13 5
13. Solve
14. Solve tan (x) = −
c
2007-2009 MathHands.com
math hands
12 5
pg. 7
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
15. Find all solutions cos x = 0 16. Find all solutions cos θ = −1 17. Find all solutions cos θ = 1 18. Find all solutions cos θ = 1/2 19. Find all solutions cos θ = .25 20. Find all solutions cos θ = −2 21. Find all solutions cos x = 1.01 22. Find all solutions cos x = .999 23. Find all solutions cos x =
−1 2
24. Find all solutions sin x = 0 25. Find all solutions sin θ = −1 26. Find all solutions sin θ = 1 27. Find all solutions sin θ = 1/2 28. Find all solutions sin θ = .25 29. Find all solutions sin θ = −2 30. Find all solutions sin x = 1.01 31. Find all solutions sin x = .999 32. Find all solutions sin x =
c
2007-2009 MathHands.com
math hands
−1 2
pg. 8
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
33. Find all solutions csc x = −2 34. Find all solutions csc x = 3 35. Find all solutions tan x = 1 36. Find all solutions tan x = −1 37. Find all solutions tan x = 5 38. Find all solutions sin x = 5 39. Find all solutions
√ − 3 sin x = 2
40. Find all solutions sin x =
√
5 6
41. Find all solutions cos x = .123 42. (extra fun points) Find all solutions cos x = sin x
c
2007-2009 MathHands.com
math hands
pg. 9
MathHands.com M´ arquez
Solving Trig Equations: The Easy Ones Main Idea We are now ready to discuss the solving of trigonometric equations. Recall that, generally speaking, identities are equations which hold true for all values. Today, we consider equations that, generally speaking, are only true for some values of x (or θ, or any variable/s). Our goal is to determine exactly for which values the equation holds true. This is the general meaning of solving equations. For example, consider the equation, x2 = 1 Clearly, this equation is not true for all real values of x. Nevertheless, we can solve it by taking careful steps:
x2 = 1
(given)
x −1=0 (x − 1)(x + 1) = 0
(algebra) (factor, algebra)
2
x−1=0 or x+1=0 x=1 or x = −1
(Zero Factor Theorem) (algebra)
Thus, this equation has two real solutions. Let us call the two solutions x1 and x2 . Then the solutions are x1 = 1 and x2 = −1
This is a classic polynomial equation, hopefully one you find familiar from previous studies. Here, we solve equations that involve the variables on the trigonometric functions, thus the description trig equations. We have actually already taken a first look at such equations. Recall we have seen equations such as sin(x) =
3 4
when we first introduced the inverse trig function, sin−1 r, or arcsin r. This was particularly useful in solving right triangles when a ratio was known, such as 3/4, and the angle was sought. At that point, we used a calculator to compute/estimate the arcsine of 3/4, in degrees mode, as sin−1 43 ≈ 48.6◦ . At that point we also talked about the domain restriction and the fact that the domain for sin−1 is only real rations, r where −1 ≤ r ≤ 1 while the codomain is only angles θ where −90◦ ≤ θ90◦ . We claimed we would come back and revisit such equations to solve completely. Well.. the time is now and the place is here. We now have all the necessary time and tools to solve, completely, correctly, and honestly, trig equations such as sin(x) =
3 4
The general strategy we will employ is to graph both sides of the equation. In other words, we will graph y = sin x. We will then graph on the same plane, y = 34 . We will then see if and where these two graphs intersect. They will intersect exactly where sin(x) = 34 . At that point, we simply have to point to the solutions and write them down. This last step may take some getting used to because, unlike x2 = 1, trig equations may have infinite many solutions.... talk is cheap.. let us solve!
c
2007-2009 MathHands.com
math hands
pg. 1
Trigonometry Sec. 01 notes
MathHands.com M´ arquez
Example Solve sin (x) =
3 4
Solution: First we graph y = 3/4 then we graph y = sin x, then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them.
2.5 2.0 y=
3 4
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.59
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0 -1.5 -2.0 -2.5
Now, we need to describe them. To do so, it may be sometimes helpful to separate all the solutions into two different ’buckets’. On one bucket we place all the ’uphill solutions’, namely the ones that occur as the sine graph increases.
c
2007-2009 MathHands.com
math hands
pg. 2
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y=
3 4
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.59
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0 -1.5 ≈ −671.41◦
360◦
≈ −311.2◦
360◦ -2.0
≈ 48◦
360◦
≈ 408◦
360◦
-2.5
Of these, the first one is determined, when the ratio is not a famous one such as 3/4, by using a calculator to estimate sin−1 (3/4) ≈ 48.59◦. This solution is the first one that occurs, right of the origin. we could call this, our fist solution, x1 , the next one occurs exactly 360◦ right of that one, x2 ≈ 408.59◦ and so on.. While on the left side, we find a solution at x−1 ≈ 48.59◦ − 350◦ = −311.41. In other words, the uphill solutions are given by a clear pattern: . . . , −311.41◦, 48.59◦, 408.59◦, . . . In fact, a more descriptive and concise way to describe such sequence is to use a closed formula that describes the pattern. Namely, we can describe all such uphill solutions by a formula. The formula would have to say that we can start with 48.59◦ (or any other uphill solution) then proceed to add or subtract any whole number multiple of 360◦ . If we allow k to be any integer [we write k ∈ Z], positive or negative, this can be accomplished by the following formula: xk ≈ 48.59◦ + k360◦
for k ∈ Z
Indeed, this is a common and effective way to describe the uphill solutions. Note one could also use radians, xk ≈ 0.848 + k2π
for k ∈ Z
This sets up a correspondence, for every integer value k, xk is an uphill solution, and every uphill solution corresponds to exactly one value of k. It should be noted that sometimes, authors leave out the k, and it is understood that each value of k will produce a corresponding value solution x, thus the following notation is common: x ≈ 48.59◦ + k360◦
for k ∈ Z
Of course, this is only half of the solutions, now we turn our attention to the downhill solutions, and proceed in a similar manner.
c
2007-2009 MathHands.com
math hands
pg. 3
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y=
3 4
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.59
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0 -1.5 ≈ −671.41◦
≈ −311.2◦
360◦
360◦ -2.0
≈ −588
◦
360
◦
≈ −228
◦
≈ 48◦
360◦ -2.5
360◦
≈ 132◦
≈ 408◦ 360◦
≈ 492◦
360◦ 360◦
We use the symmetry of the graph to determine one of the solutions. For example, the first solution right of the origin is ≈ 180◦ − 48.59◦, and in this case as well, all the downhill solutions are equally spaced 360◦ apart, thus we can write a closed form for all the downhill solutions: xk ≈ 131.41◦ + k360◦
for k ∈ Z
We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 48.59◦ + k360◦
for k ∈ Z
xk ≈ 131.41◦ + k360◦
for k ∈ Z
OR
Example Solve sin (x) =
1 3
Solution: First we graph y = 0.333 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (0.333) ≈ 19.471◦. This solution is the only one provided by the sin−1 function.
c
2007-2009 MathHands.com
math hands
pg. 4
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y=
1 3
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.471
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0
≈ −700.529◦
360◦
≈ −340.30◦
360◦
-1.5 ≈ 19.28◦
360◦
-2.0 ≈ −559
◦
360
◦
≈ −199
◦
360◦ -2.5
≈ 161◦
≈ 380.8◦ 360◦
360◦ ≈ 521◦
360◦
We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 19.471◦ + k360◦
for k ∈ Z
xk ≈ 160.529◦ + k360◦
for k ∈ Z
OR
Example Solve sin (x) = −
3 5
Solution: First we graph y = −0.6 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (−0.6) ≈ −36.87◦. This solution is the only one provided by the sin−1 function.
c
2007-2009 MathHands.com
math hands
pg. 5
Trigonometry Sec. 01 notes
MathHands.com M´ arquez 2.5 2.0 y= −
3 5
y = sin(x)
1.5 1.0 .5
-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦
.13
90◦
180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦
-.5 -1.0
≈ −756.87◦
360◦
≈ −396.8◦
-1.5 ≈ −36◦
360◦
360◦
-2.0 ≈ −503
◦
360
◦
≈ −143
360◦
◦
-2.5
≈ 217◦
≈ 324◦ 360◦
360◦ ≈ 577◦
360◦
We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ −36.87◦ + k360◦
for k ∈ Z
xk ≈ 216.87◦ + k360◦
for k ∈ Z
OR
c
2007-2009 MathHands.com
math hands
pg. 6
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solving Trig Equations: The Easy Ones
1. Solve sin (x) =
2 5
sin (x) =
4 5
sin (x) =
6 7
2. Solve
3. Solve
4. Solve
5 11
sin (x) = − 5. Solve cos (x) =
7 10
6. Solve
7 10
cos (x) = − 7. Solve cos (x) =
3 5
cos (x) =
9 10
8. Solve
9. Solve cos (x) = −
8 15
cos (x) = −
3 4
tan (x) = −
3 4
10. Solve
11. Solve
12. Solve tan (x) =
3 5
tan (x) =
13 5
13. Solve
14. Solve tan (x) = −
c
2007-2009 MathHands.com
math hands
12 5
pg. 7
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
15. Find all solutions cos x = 0 16. Find all solutions cos θ = −1 17. Find all solutions cos θ = 1 18. Find all solutions cos θ = 1/2 19. Find all solutions cos θ = .25 20. Find all solutions cos θ = −2 21. Find all solutions cos x = 1.01 22. Find all solutions cos x = .999 23. Find all solutions cos x =
−1 2
24. Find all solutions sin x = 0 25. Find all solutions sin θ = −1 26. Find all solutions sin θ = 1 27. Find all solutions sin θ = 1/2 28. Find all solutions sin θ = .25 29. Find all solutions sin θ = −2 30. Find all solutions sin x = 1.01 31. Find all solutions sin x = .999 32. Find all solutions sin x =
c
2007-2009 MathHands.com
math hands
−1 2
pg. 8
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
33. Find all solutions csc x = −2 34. Find all solutions csc x = 3 35. Find all solutions tan x = 1 36. Find all solutions tan x = −1 37. Find all solutions tan x = 5 38. Find all solutions sin x = 5 39. Find all solutions
√ − 3 sin x = 2
40. Find all solutions sin x =
√
5 6
41. Find all solutions cos x = .123 42. (extra fun points) Find all solutions cos x = sin x
c
2007-2009 MathHands.com
math hands
pg. 9
