Download 7.5 Trigonometric Equations One frequently has to solve equations

7.5
Trigonometric Equations
One frequently has to solve equations involving trig functions. Sometimes the values of x
you look at are restricted, while others you are asked to find all the values of x that make
a given equation true. In the latter case, there are usually an infinite number of solutions
whose form depends on the period of the trig functions involved.
Examples:
1. Solve
(a) tan2 (x) − 3 = 0
(b) sin(x) = cos(x)
(c) 1 + sin x = 2 cos2 (x)
(d) sin 2x − cos x = 0
(e) cos(x) + 1 = sin x with t ∈ [0, 2π]
solution:
√
(a) tan2 (x) − 3 = 0 ⇒ tan(x) = ± 3.
√
tan(x) = 3 when x = π3 . But tan has period π, so
√
π
tan(x) = 3 ⇒ x = + kπ k ∈ Z
3
Likewise,
√
π
tan(x) = − 3 ⇒ x = − + kπ k ∈ Z
3
Thus the full set of solutions is
n π
o
π
S = − + kπ, x = + kπ | k ∈ Z
3
3
(b) If sin(x) = cos(x), cos(x) 6= 0 because sin and cos are not 0 in the same places.
Thus we can divide both sides by cos and get
sin(x)
= 1 ⇒ tan(x) = 1
cos(x)
This happens when x = π4 . Once again, tan has a period of π so
π
x = + kπ k ∈ Z
4
(c) We can get the equation 1 + sin x = 2 cos2 (x) into an easier form to deal with by
subbing in 1 − sin2 x for cos2 x:
1 + sin x = 2 cos2 (x)
1 + sin x = 2 − 2 sin2 x
2 sin2 x + sin x − 1 = 0
This is a quadratic in sin x that factors as
(2 sin x − 1)(sin x + 1) = 0
Which means either 2 sin x − 1 = 0 or sin x + 1 = 0. The first gives us that
sin x = 12 , and the second gives sin x = −1.
• sin x = −1 ⇒ x =
• sin x =
1
2
⇒x=
π
6
3π
2
+ 2kπ, k ∈ Z
+ 2kπ, k ∈ Z or x =
5π
6
+ 2kπ, k ∈ Z
(d) We use a double angle identity on sin 2x − cos x = 0 to get
2 sin x cos x − cos x = 0
cos x(2 sin x − 1) = 0
Hence either
• cos x = 0 ⇒ x =
• sin x =
1
2
⇒x=
π
2
π
6
+ 2kπ, k ∈ Z or x =
+ 2kπ, k ∈ Z or x =
3π
2
5π
6
+ 2kπ, k ∈ Z.
+ 2kπ, k ∈ Z.
(e) Here we have to be a little trickier and square both sides
cos x + 1 = sin x
(cos x + 1)2 = 1 − cos2 x
cos2 x + 2 cos x + 1 = 1 − cos2 x
2 cos2 x + 2 cos x = 0
2 cos x(cos x − 1) = 0
Hence either cos x = 0 or cos x = −1. Between 0 and 2π, the first only happens
at π2 and 3π
and the second happens at −π. Hence the three solutions are
2
x=
π 3π
, ,π
2 2
2. Consider the equation 2 sin(3x) − 1 = 0.
(a) Find all the solutions to the equation.
(b) Find all the solutions to the equation in the interval [0, 2π).
solution:
(a) The equation rearranges to sin(3x) = 12 . As we’ve seen before, this means that
π
5π
+ 2kπ, k ∈ Z or 3x =
+ 2kπ, k ∈ Z.
6
6
π
2kπ
5π 2kπ
x=
+
, k ∈ Z or x =
+
, k ∈ Z.
18
3
18
3
3x =
(b) We solve the inequalities
π
2kπ
+
< 2π
18
3
1
2k
0≤
+
<2
18
3
2k
1
35
1
<2−
=
− ≤
18
3
18
18
1
35
− ≤ 2k <
6
6
35
1
≈ 2.91
− ≤k<
12
12
The ks in this range are k = 0, k = 1 and k = 2. Hence
0≤
x=
π 13π 25π
,
,
18 18 18
For the other solutions we have
0≤
5π 2kπ
+
< 2π
18
3
5
2k
+
<2
18
3
5
2k
5
31
− ≤
<2−
=
18
3
18
18
31
5
− ≤ 2k <
6
6
31
5
≈ 2.58
− ≤k<
12
12
The ks in this range are k = 0, k = 1 and k = 2. Hence
0≤
x=
3. Consider the equation
√
5π 17π 29π
,
,
18 18 18
3 tan( x2 ) − 1 = 0.
(a) Find all the solutions of the equation.
(b) Find all the solutions in the interval [0, 4π).
solution:
(a) This rearranges to tan( x2 ) =
√1
3
This gives us
x
π
= + kπ, k ∈ Z
2
6
π
x = + 2kπ, k ∈ Z
3
(b) As before, we find the ks we need:
0≤
π
+ 2kπ < 4π
3
1
+ 2k < 4
3
1
1
11
− ≤ 2k < 4 − =
3
3
3
1
11
− ≤k<
≈ 1.83
6
6
The ks in this range are k = 0 and k = 1. Hence x =
0≤
π
3
or x =
7π
.
3
4. Solve the equation tan2 x − tan x − 2 = 0.
solution: This is a quadratic in tan x, so
tan2 x − tan x − 2 = 0
(tan x − 2)(tan x + 1) = 0
So tan x = 2 or tan x = −1. There isn’t a convenient angle with tan x = 2, but we can
use tan−1 to write
x = tan−1 (2) + kπ, k ∈ Z
The second one of course gives us
x=−
π
+ kπ, k ∈ Z
4
5. Solve the equation 3 sin θ − 2 = 0.
solution: This rearranges to sin θ = 32 . Once again there is no easy angle that gives
us sin θ = 23 . We know that sin is positive in the first and second quadrants, and thus
the two solutions in [0, 2π) are
2
2
−1
−1
sin
and π − sin
3
3
Hence the full solution is
−1
θ = sin
2
2
−1
+ 2kπ or θ = π − sin
+ 2kπ
3
3