A2HCh0703 Solving Trigonometric Equations
Homework and Reading
Read p551– 557
(1) HW p558 #1– 43 odd
(2) HW p559 #45 – 61odd 73, 74
Goal
Solving Trigonometric Equations
p1
Students use standard algebraic techniques to solve trigonometric equations, solve
trigonometric equations of quadratic type, solve trigonometric equations involving multiple
angles, and use inverse trigonometric functions to solve trigonometric equations.
Solving Trigonometric Equations
To solve a trigonometric equation, try to
isolate the trigonometic function on one side,
then find the solutions.
Using a let statement can sometimes help
with more complicated equations.
Use know trigonometric values to find
solutions.
Solving Trigonometric Equations
Example
Solving Trigonometric Equations
Solve sin x + 1 = ! sin x
Example
Solve 1 ! 2 cos x = 0
!
3
1 ! 2 cos x = 0
isolate cos x
! 2 cos x = !1
subtract 1
1
cos x =
divide ! 2, find x
2
"
x = + 2" n
3
5"
or
+ 2" n and cotermial
3
There are infinitely many solutions, use the
period for the function. + 2" n
5!
3
Example
Example
Solve tan 2 x ! 3 = 0
tan 2 x ! 3 = 0
tan 2 x = 3
2!
3
!
3
isolate tan x
add 3
tan x = ± 3
take sqr root
"
x = + "n
3
2"
or
+ " n and cotermial
3
There are infinitely many solutions, use the
period for the function. + " n
Equations of Quadratic Type
Solve 2 cos x + cos x ! 1 = 0 on the interval [ 0, 2" )
2u 2 + u ! 1 = 0
( 2u ! 1) (u + 1) = 0
2u ! 1 = 0 u + 1 = 0
1
u = !1
u=
2 cos x = !1
1
cos x =
2
" 5" x = "
x= ,
3 3
sec x csc x = csc x
sec x csc x ! csc x = 0
isolate sec x & csc x
subtract csc x
csc x ( sec x ! 1) = 0
factor csc x
csc x = 0 sec x ! 1 = 0
sec x = 1
two solutions
Undefined
x = 2" n
Solve 2 cos 2 ! + 3sin ! " 3 = 0 on the interval [ 0, 2# )
2
let u = cos x
Solve sec x csc x = csc x
Example
Equations of Quadratic Type
Example
sin x + 1 = ! sin x
isolate sin x
2 sin x + 1 = 0
add sin x
7!
6
1
sin x = !
solve
2
7"
x=
+ 2" n
6
11"
or
+ 2" n and cotermial
6
There are infinitely many solutions, use the
period for the function. + 2" n
Use let statement u for cos x
substitute u for cos x
factor
(
)
2 1 " sin 2 ! + 3sin ! " 3 = 0
(
)
let u = sin !
2 1 " u 2 + 3u " 3 = 0
2 " 2u 2 + 3u " 3 = 0
"2u 2 + 3u " 1 = 0
2u 2 " 3u + 1 = 0
( 2u " 1) (u " 1) = 0
solve each factor, replace cos x
2u " 1 = 0
u "1= 0
1
u =1
u=
sin ! = 1
2
1
sin ! =
2
#
# 5#
!=
!= ,
2
6 6
Use identities for cos !
2
Use let statement u for sin !
substitute u for sin !
distribute
Simplify
factor a " 1
factor
solve each factor, replace sin !
11!
6
A2HCh0703 Solving Trigonometric Equations
Examples & Try
Solve sec x + 1 = tan x on the interval [ 0, 2! )
Example
(sec x + 1) = tan x
(sec x + 1)2 = sec 2 x " 1
2
2
let u = sec x
(u + 1)
2
= u "1
u + 2u + 1 = u " 1
2u = "2
u = "1
sec x = "1
x=!
2
Functions Involving Multiple Angles
square both sides
use the identity
Use let statement u for sec x
2
2
substitute u for sec x
multiply
Simplify
simplify
replace u with sec x
solve
Solve 2 sin 2 2x = 1 on the interval [ 0, 2! )
Solve
sin 2! + 1 = 0 on the interval [ 0, 2" )
Wo2rk
sin 2 2x =
1
7"
!sin
= 2!+ "=n #
11"
isolate
12
! =sin 2!+ " n
2
n
12
!
n
7"
!
7
"
11
"
7
"
0
+!"=
11
"
0
(
=
)
2
or
solve
11"
12
0
+
6 712" 126
12 " ( 0 ) = 12
7"
"
19
1
"
11"
11" 12" 23"
12
1
7"+ " (1) = 12 + 12 =11
+
12"
12 " (1) = 12 + 12 =
2!2= 7" + "+( 22" n7" 224
12
) = + ! "== 31" + 22" n11" +
11
" 24" 35"
126
12
6
12
12
12 " ( 2 ) = 12 + 12 =
7"
12
11
"
! = 7" +19"" n
!=
+ " n 11" solve
each
! =12 ,
23"
12
!=
,
712" 12
19"
11" 23" 12 12
!=
,
!=
,
12 12
12 12
Solutions :
!
7" 11" 19" 23"
,
,
,
12 12 12 12
!
4x = + ! n
4
! !n
x=
+
16 4
! 5!
"= ,
,...
16 16
Solutions :
Using Inverse Functions
!
5!
or
4
4
solve
5!
+ !n
4
5! ! n
x=
+
16
4
5! 9!
"=
,
,...
16 16
4x =
! 5! 9! 13! 17! 21! 25! 29!
,
,
,
,
,
,
,
16 16 16 16 16 16 16 16
Using Inverse Functions
Example
Solve sec 2 x ! 3sec x ! 10 = 0
let u = sec x
Use let statement u for sec x
u 2 ! 3u ! 10 = 0
(u ! 5 )(u + 2 ) = 0
u!5=0
u=5
sec x = 5
substitute u for sec x
factor
u+2=0
u = !2
sec x = !2
x = sec !1 5 + 2" n x = sec !1 ( !2 ) + 2" n
solve and replace u
1
2
isolate sin 2x
2
2
1
2
2
!
3!
2x = or
4
4
sin 2x =
!
2x = + ! n
4
! !n
x= +
8 2
! 5! 9! 13!
"= ,
,
,
8 8 8
8
Solutions :
Try : Solve tan 4x = 1 on the interval [ 0, 2! )
4x =
Try : Solve 3cot x ! 1 = 0
1
2
cot 2 x =
isolate cot x
3
1
cot x = ±
sqr root
3
tan x = ± 3
use identity
"
x = + "n
3
2"
or x =
+ "n
3
Example
Functions Involving Multiple Angles
Example
!
p2
2
sqr root
0
!
2!
-1
solve
-2
3!
+ !n
4
3! ! n
x=
+
8
2
3! 7! 11! 15!
"=
,
,
,
8 8
8
8
2x =
period is !
since we had
2
sin 2x, no
negatives.
! 3! 5! 7! 9! 11! 13! 15!
,
,
,
, ,
,
,
8 8 8 8 8
8
8
8
Try : Solve sin 2 3x ! 2 sin x + 1 = 0 on the interval [ 0, 2" )
let u = sin x
u 2 ! 2u + 1 = 0
(u ! 1) (u ! 1) = 0
u !1= 0
sin 3x ! 1 = 0
sin 3x = 1
"
3x = + 2" n
2
" 2" n
x= +
6
3
" 5" 9"
x= ,
,
6 6 6
Use let statement u for sin x
substitute u for sin x
factor
replace u with sin x
solve
solve
solve