November 26, 2014
Section 5.3 Solving Trigonometric Equations
Section Objectives: Use standard algebraic techniques and inverse
trig. functions to solve trig. equations.
Ex1: Solve 1 - 2x = 0
To solve trig. equations
use algebraic techniques
⇒collecting like terms,
factoring,...
1 - 2cos x = 0.
General Solut
x = π/3 + 2πn
5π/3 + 2π
GOAL: Isolate the trig. function!
?
solutions
y
n
a
m
How
***Remember, you can check solutions graphically!***
ber
m
e
Rem
!
Unit Circle
(cos x, sin x)
November 26, 2014
Ex2: Solve sin x + 1 = -sin x
2sin x + 1 = 0
sin x = -1/2
x = 7π/6 + 2πn or x =
Ex3: Solve tan2 x - 3 = 0
tan 2x = 3
tan x = ±√3
x = π/3 + πn or x
Ex4: Solve sec x csc x = csc x
t.,
rig. fc
t
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sec x csc x - csc x =
csc x(sec x - 1) =
csc x = 0 or sec x
Undef.
Quadratic - type Equations?
!
Yes
Ex5: Solve the following on the interval [0, 2π).
a) 2cos 2 x + cos x - 1 = 0
b) 2cos 2 x + 3sin x - 3 = 0
c) cos x + 1 = sin x
Need 1 trig. function.
2cos 2 x + cos x - 1 = 0
(2cos x - 1)(cos x + 1) = 0
2cos x - 1 = 0 or cos x + 1 = 0
cos x = 1/2 or cos x = -1
x = π/3, 5π/3 or x = π
a)
2cos2 x + 3sin x - 3 = 0
2(1 - sin2 x) + 3sin x - 3 = 0
2sin2 x - 3sin x + 1 = 0
(2sin x - 1)(sin x - 1) = 0
2sin x - 1 = 0 or sin x - 1 = 0
2 or sin x =
sinxx += 1)
1/2
c) (cos
= (sin x)2 1
x = π/6,
5π/6, or π/2
2
2
b)
cos x + 2cosx + 1 = sin x
cos2x + 2cos x + 1 - sin2x = 0
2cos2x + 2cos x = 0
2cos x (cos x + 1) = 0
2cos x = 0
cos x + 1 = 0
cos x = 0
cos x = -1
x = π/2, 3π/2
x =π
*Since squared original, check for
extraneous roots.
sec
November 26, 2014
Functions Involving Multiple angles!
Ex6: Solve the following on the interval [0, 2π).
a) 2sin2 (2t) - 1 = 0
a) 2sin2 (2t) - 1 = 0
sin2 2t = 1/2
sin 2t = ±√2/2
2t = π/4 +
2t = 3π/4
t = π/8, 3π/8
Tip: Note that since 0 ≤ t ≤ 2π, 0 ≤ 2t ≤ 4π
b) cot (x/2) + 1 = 0
b) cot (x/2) + 1 = 0
cot (x/2) = -1
x/2 = 3π
x = 3π
Tip: Note that since 0 ≤ x ≤ 2π, 0 ≤ x/2 ≤ π
Using Inverse Functions...
Ex7: Solve sec2 x – 2tan x = 4
Period for tan x = π
Interval for tan-1x = (-π/2, π/2)
sec2 x – 2 tan x – 4
1 + tan2x - 2 tan x -4
tan2x - 2 tan x - 3 = 0
(tan x - 3)(tan x + 1) = 0
tan x – 3 = 0 or tan x +
tan x = 3 or tan x =
x = tan-1 (3) + nπ
or
x = tan-1 (-1) = -π/4 + nπ