Pre-Calculus
149
Trigonometric Equations
Solve the following:
2 sin x – 1 = 0
2 sin x = 1
sin x = ½
Solve on ℜ
5π
+ 2nπ
6
6
------------------------------------------------------------------sin (x ) + 2 = − sin (x ) Solve on ℜ
x=
π
+ 2nπ and x =
2 sin (x ) = − 2
sin (x) = −
2
2
5π
7π
+ 2nπ and x =
+ 2nπ
4
4
------------------------------------------------------------------3 tan 2 x − 1 = 0 Solve on ℜ
x=
3 tan 2 x = 1
1
tan 2 x =
3
1
3
tan x = ±
=±
3
3
π
5π
x = + nπ and x =
+ nπ
6
6
------------------------------------------------------------------cot x cos2x = 2 cot x Solve on ℜ
cot x cos2x – 2 cot x = 0
cot x(cos2x – 2) = 0
cot x = 0 and cos2x = 2
cot x = 0 and cos x = ± 2 Not possible
π
x = + nπ
2
------------------------------------------------------------------2 sin2x – sin x – 1 = 0 Solve on the interval [0, 2 π )
(2 sin x + 1)(sin x – 1) = 0
sin x = -1/2 and sin x = 1
π
7π
11π
x=
x=
and
2
6
6
2 sin2x + 3 cos x – 3 = 0 Solve on ℜ
2(1 – cos2x) + 3 cos x – 3 = 0
-2 cos2x + 3 cos x – 1 = 0
2 cos2x – 3 cos x + 1 = 0
(2 cos x – 1)(cos x – 1) = 0
cos x = ½ and cos x = 1
5π
π
x = + 2nπ and
+ 2nπ and 2nπ
3
3
------------------------------------------------------------------cos x + 1 = sin x Solve on the interval [0, 2 π )
cos2x + 2 cos x + 1 = sin2x Square Both Sides
cos2x + 2 cos x + 1 = 1 – cos2x
2 cos2x + 2 cos x = 0
2 cos x(cos x + 1) = 0
cos x = 0 and cos x = -1
π
3π
x = and
and π
2
2
CAUTION: Whenever we square both sides, we introduce the possibility of extraneous answers.
Therefore we must check the solutions in the original equation.
⎛ π ⎞
⎛ π ⎞
cos⎜ ⎟ + 1 = 1, sin ⎜ ⎟ = 1 Check
⎝ 2 ⎠
⎝ 2 ⎠
⎛ 3π ⎞
⎛ 3π ⎞
cos⎜ ⎟ + 1 = 1, sin ⎜ ⎟ = −1 Does Not Check
⎝ 2 ⎠
⎝ 2 ⎠
cos(π ) + 1 = 0, sin(π ) = 0 Check
π
and π
2
------------------------------------------------------------------2 cos 3t – 1 = 0 Solve on ℜ
cos(3t) = ½
π
5π
3t = + 2nπ and
+ 2nπ
3
3
π 2nπ
5π 2nπ
t= +
and
+
9
3
9
3
------------------------------------------------------------------3 tan2x = 5 sec x - 1 Solve on ℜ
3(sec2x - 1) = 5 sec x - 1
3 sec2x - 3 = 5 sec x - 1
3 sec2x - 5 sec x - 2 = 0
(3 sec x + 1)(sec x - 2) = 0
1
sec x = −
and sec x = 2
3
The only valid statement is sec x = 2
π
5π
x = + 2nπ and x =
+ 2nπ
3
3
x=
sec2x - 2 tan x = 4 Solve on ℜ
tan2x + 1 – 2 tan x – 4 = 0
tan2x – 2 tan x – 3 = 0
(tan x - 3)(tan x + 1) = 0
tan x = 3 and tan x = -1
3π
x = arctan 3 + n π and x =
+ nπ
4
------------------------------------------------------------------sin x = 6 cos2x Solve on ℜ
sin x = 6(1 - sin2x)
sin x = 6 - 6 sin2x
6 sin2x + sin x - 6 = 0
In this case factoring is not possible, so we must use the Quadratic Formula
a = 6, b = 1, c = -6
− 1 ± 1 − 4(6)(−6)
− 1± 145
=
sin x =
12
12
− 1− 145
sin x =
is not valid because it is less than -1
12
− 1 + 145
Therefore sin x =
12
⎛ − 1 + 145 ⎞
⎛ − 1 + 145 ⎞
π
⎟ + 2n π and x =
⎟ + 2n π
x = arcsin ⎜
- arcsin ⎜
⎜
⎟
⎜
⎟
2
12
12
⎝
⎠
⎝
⎠
-------------------------------------------------------------------
Pre-Calculus Assignment 149 Monday January 4, 2016 Hour
Name
Find all solutions of the equation in the interval [0, 2π) algebraically.
1.
2 sin x + 1 = 0
2.
tan2x - 1 = 0
3.
2 sin2x = 2 + cos x
4.
sec x csc x = 2 csc x
5.
sec x + tan x = 1
6.
sin2x + cos x + 1 = 0
7.
2 cos2x + cos x - 1 = 0
Pre-Calculus Assignment 149 Monday January 4, 2016 Hour
Name
Ewell-Key
Find all solutions of the equation in the interval [0, 2π) algebraically.
1.
2 sin x + 1 = 0
1
7π 11π
6
sin x = - 2 x = 6
2.
tan2x - 1 = 0
tan x = ± 1
3.
2 sin2x = 2 + cos x
π 3π 5π 7π
x =4 4 4 4
2 - 2 cos2x = 2 + cos x
2 cos2x + cos x = 0
1
π 3π 2π 4π
cos x(2 cos x + 1) = 0 → cos x = 0 cos x = - 2 x = 2 2 3 3
4.
sec x csc x = 2 csc x
sec x csc x - 2 csc x = 0
csc x(sec x - 2) = 0 → csc x = 0 sec x = 2
π 5π
3
x=3
5.
sec x + tan x = 1
1
cos x
sin x
cos x
+ cos x - cos x = 0
1 + 2 sin x + sin2x = cos2x
2 sin2x + 2 sin x = 0
6.
sin2x + cos x + 1 = 0
1 + sin x - cos x = 0
1 + 2 sin x + sin2x = 1 - sin2x
sin x(sin x + 1) = 0 x = 0 π
1 - cos2x + cos x + 1 = 0
cos2x - cos x - 2 = 0
(cos x - 2)(cos x + 1) = 0 x = π
7.
1 + sin x = cos x
2 cos2x + cos x - 1 = 0
(2 cos x - 1)(cos x + 1) = 0
π 5π
x=3 3 π
!!
!
x=0 .