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Trigonometric Equations with
Multiple Angles
The diagram shows where the various ratios are positive
Note also that values
repeat every 360o
90°
180 -
S
A
180°
180 +
0°,360°
C
T
270°
360 -
Solve cos 3x = – 1/√2
for values of x such that 0 < x < π
First of all consider the range of values
If 0 < x < π then 0 < 3x < 3π (540o)
o =to
o
Work in Remember
degrees
then
3π
change
=
3
×
180
back
540
radians
later
Now find BASE angle
cos -1 (1/√2 ) = 45o
π/
4
180 –

S
A
Since cosine negative
T
C
o
o
o

3x = 135 , 225 495
180 +
Now change back to radians using multiples of π/4
3x = 3π/4 , 5π/4 , 11π/4
So x = 3π/12 , 5π/12 , 11π/12
Solve sin 2x = 1/2
for values of x such that 0 < x < 2π
First of all consider the range of values
If 0 < x < 2π then 0 < 2x < 4π (720o)
o =to
o
Work in Remember
degreesNow
then
find
change
back
angle
radians
later
4π
= 4BASE
× 180
720
sin -1 (1/2 ) = 30o
π/
6
180 –

S
A
Since sine positive
T
C
o
o
o
o
2x = 30 , 150 390 , 510
Now change back to radians using multiples of π/6
2x = π/6 , 5π/6 , 13π/6 ,17π/6
So x = π/12 , 5π/12 , 13π/12 , 17π/12
Solve tan 4x = √3
for values of x such that 0 < x < π
First of all consider the range of values
If 0 < x < π then 0 < 4x < 4π (720o)
o =to720
o
Work in degrees
Remember
Now
then
find
4πchange
=BASE
4 × 180
back
angle
radians
later
tan -1 (√3 ) = 60o
π/
3
S
A
Since tangent positive
T
C
180 + 
o
o
o
o
4x = 60 , 240 420 , 600
Now change back to radians using multiples of π/3
4x = π/3 , 4π/3 , 7π/3 ,10π/3
So x = π/12 , π/3 , 7π/12 , 5π/6
Solve 2sin (2x – π/6) = √3
for values of x such that 0 < x < 2π
First of all consider the range of values
If 0 < x < 2π then 0 < 2x – π/6 < 4π – π/6) (690o)
π/ change
o – to
o = 690olater
Work
Remember
in degrees
Now
2πthen
–find
=
BASE
2
×
180
back
angle
30
radians
6
sin -1 (√3/2 ) = 60o
2x –
π/
6
=
60o
,
π/
120o
180 -
3
420o , 480o
S

A
T
C
Now change back to radians using multiples of π/3
2x – π/6 = π/3 , 2π/3 , 7π/3 ,8π/3
2x =
3π/
6
,
5π/
6
,
15π/
6
,17π/
6
So x = π/4 , 5π/12 , 5π/6 , 17π/12