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Higher
Higher Maths
Connection between Radians / degrees
& Exact values
Solving Basic Trig Equations
Solving Harder Trig Equations
Solving Trig Equation by Substitution
Solving Wave Function
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Radians
Outcome 3
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Higher
Radian measure is an alternative to degrees
and is based upon the ratio of
arc Length
radius
L
θ
θ- theta
(angle at the centre)
r
Circumference  2 r
Circumference
 2
r
So, full circle
360o 2π radians
Radians
Copy Table
90o  π
2
π
o
60 
3
π
o
45 
4
30o
π

6
360o 
2π
270o  3π
2
2π
o
120 
3
3π
o
135 
4
240o
5π

6
210o
150o
Demo
180o 
4π

3
5π
o
225 
4
7π

6
π
5π

3
7π
o
315 
4
300o
330o
11π

6
Converting
Outcome 3
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Higher
For any values
÷180 then
X
π
degrees
radians
÷ π then x 180
Converting
Outcome 3
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Higher
Ex1
72o =
Ex2
330o = 330/180 X π = 11 π /6
Ex3
2π /9 =2π /9 ÷ π x 180o =
Ex4
72/
180
X
π = 2π /5
2/
9
23π/18 = 23π /18 ÷ π x 180o =
X
180o = 40o
23/
18
X
180o = 230o
Exact Values
Outcome 3
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Higher
Some special values of Sin, Cos and Tan are useful left
as fractions, We call these exact values
2
60º
60º
2
2
3
60º
30º 2
1
This triangle will provide exact values for
sin, cos and tan 30º and 60º
60º
Exact Values
Outcome 3
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Higher
x
Sin xº
Cos xº
Tan xº
0º
0
30º
45º
60º
½
3
3
½
2
1
3
90º
2
3

Exact Values
Outcome 3
Higher
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Consider the square with sides 1 unit
1
1
2
45º
1
45º
1
We are now in a position to calculate
exact values for sin, cos and tan of 45o
Exact Values
Outcome 3
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Higher
x
Sin xº
Cos xº
Tan xº
0º
0
30º
45º
60º
½
1
2
3
3
2
1
2
½
1
3
1
3
90º
2
Undefined
Solving Trig Equations
Outcome 3
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Higher
Sin +ve
All +ve
180o - xo
1
2
3
4
created by Mr. Lafferty
180o + xo
360o - xo
Tan +ve
Cos +ve

o
Solving Trig Equations
90
Graphically what
2
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Higher
to
Outcome 3are we trying
A
S
o
 solve
180
Example 1 Type 1:
T C
Solving the equation sin
xo
= 0.5 in the range
0o
0o
o
270
o
to 360
3
2
sin xo = (0.5)
xo = sin-1(0.5)
1st Q xo = 30o
2nd Q xo = 150o
1
2
3
4
created by Mr. Lafferty
(180o – 30o = 150o)

o
Solving Trig Equations
90
Graphically what
2
to A
Outcome 3are we trying
S
 solve
180o
T C
Higher
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Example 2 :
Solving the equation cos
xo
0o
o
270
o
- 0.625 = 0 in the range 0 to 360o
3
2
cos xo = 0.625
xo = cos -1 (0.625)
1st Q xo = 51.3o
2nd Q xo = 308.7o
1
2
3
4
created by Mr. Lafferty
(360o - 53.1o = 308.7o)

o
Solving Trig Equations
90
Graphically what
2
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Higher
to
Outcome 3are we trying
A
S
o
 solve
180
Example 6 Type 2 :
T C
Solving the equation
cos2x
= 1 in the range
0o
to
360o
270o
3
2
cos2 xo = 1
cos xo = ± 1
cos xo = 1
xo = 0o and 360o
cos xo = -1
created by Mr. Lafferty
xo = 180o
0o
Maths4Scotland
Higher
Find the exact solutions of 4sin2 x = 1, 0  x  2
Rearrange
sin 2 x 
1
4
Take square roots
sin x  
1
2
Find acute x
acute
x 

+ and – from the square root requires all 4 quadrants
Determine quadrants for sin x
S
6
A

T
C
5
7 11
x  ,
,
,
6
6
6
6
Hint
Previous
Table of exact values
Quit
Quit
Next

o
Solving Trig Equations
90
Graphically what
2
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Higher
to
Outcome 3are we trying
A
S
o
 solve
180
Example 4 Type 3 :
T C
Solving the equation sin
2xo
+ 0.6 = 0 in the range
0o
0o
o
270
to 360o
3
2
sin 2xo = (-0.6)
2xo = sin-1(0.6)
2xo = 37o ( always 1st Q First)
2xo = 217o , 323o
o , 683o ......
577
÷2
created by Mr. Lafferty
xo =
108.5o , 161.5o
288.5o , 341.5o
Solving Trigonometric Equations
Higher
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Example:
Solve 2cos3x  1  0
Step 1: Re-Arrange
(0  x  360o )
Step 2: consider what solutions
are expected

2
2cos3x  1  0
2cos3x  1
1
cos3x 
2
90o

180
o
S
A
T
C
270o
3
2
0o
3x means
Solving Trigonometric Equations 3 rows
Higher
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Step 3: Solve the equation
1
cos3x 
2
1st Q 3x = 60o 4th Q
420o
780o
1
3 x  cos    60o
2
1
300o
660o
1020o
x = 20o
140o
100o
220o
260o
340o
Solving Trigonometric Equations
Higher
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Graphical solution for
1
cos3x 
2
Solving Trigonometric Equations
Higher
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Example:
Solve 1  2 sin 6t  0
Step 1: Re-Arrange
(0  t  180o )
Step 2: consider what solutions
are expected
1
sin 6t 
2

2
90o
sin 6t is negative so solutions in
the third and fourth quadrants
Since
0  t  180o
has 2 solutions
x6
x6
Then

0  6t  1080o
has 12 solutions
180
o
S
A
T
C
270o
3
2
0o
6x means 6 rows
Solving Trigonometric but
Equations
only over 180o
so 3 rows
Higher
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Step 3: Solve the equation
1
sin 6t 
2
3rd Q
 1 
6t  sin 1 

2


 1 
o
st
sin 1 
  45 always 1 Quad first
 2
6t = 225o
4th Q = 315o
x = 37.5o
52.5o
585o
675o
97.5o
112.5o
945o
1035o
157.5o
172.5o
Solving Trigonometric Equations
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Higher
Graphical solution for
1
sin 6t 
2

90o Solving
2

Higher
A
S
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180o
0o
Example
T C5 Type 3 :
Trig Equations
Graphically what
Outcome 3are we trying to
solve
o
270
Solving the equation 2sin (2xo - 30o) - √3 = 0 in the range 0o to 360o
3
2
2sin (2x - 30o) = √3
sin (2x - 30o) = √3 ÷ 2
(2x - 30o) = sin-1(√3 ÷ 2)
(2xo - 30o) = 60o 120o
420o 480o
2xo = 90o 150o
450o 510o
÷2
created by Mr. Lafferty
xo = 45o 75o
225o 255o
Trig Equations
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Higher
Example
Find the value of x that minimises the expression
cosxcos32 + sinxsin32
Using rule 2(b) we get
cosxcos32 + sinxsin32 = cos(x – 32)
cos graph is roller-coaster
min value is -1 when angle = 180
ie
x – 32o = 180o
ie
x = 212o
ALWAYS
work out
Quad 1
first
Higher Example 5
Trig Equations
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Solve
sinxcos30 + cosxsin30 = -0.966
where 0o < x < 360o
By rule 1a
sinxcos30 + cosxsin30 = sin(x + 30)
sin(x + 30) = -0.966
1st Q sin-1 (0.966) = 75
3rd Q
S
180-xo
A
xo
o
180+xo 360-x
T
C
(x + 30o) = 255o
4th Q
x + 30o = 285o
x = 225o
x = 255o
The solution is to be in radians – but
workTrigonometric
in degrees and convert
at the
Solving
Equations
end.
Higher
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Example:

Solve 2sin(2 x  )  1
3
Step 1: Re-Arrange
(0  x  2 )
Step 2: consider what
solutions are expected

2
1
sin(2 x  60 ) 
2
o
(2x – 60o ) = sin-1(1/2)
90o

180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Higher
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Step 3: Solve the equation
1
sin(2 x  60 ) 
2
o
1
2 x  60  sin  
2
1
o
1st Q
30o
2nd Q
150°
2x - 60° = 30o 150o
390o 510o
2x = 90o
210o
450o
570o
x = 45o
225o

4
105o
285o
7
12
5
4
19
12
Solving Trigonometric Equations
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Higher
Graphical solution for
1
sin(2 x  60 ) 
2
o


2
o
90
S A
180
0o
Example
T C7 Type 5 :
Higher
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
Solving Trig Equations
o
Outcome 3

180
Solving the equation
270o
3sin2x +32sin x - 1 = 0 in the range 0o to 360o
2
Let p = sin x
We have 3p2 + 2p - 1 = 0
Factorise (3p – 1)(p + 1) = 0
3p – 1 = 0
p = 1/3
sin x = 1/3
xo = 19.5o and 160.5o
o
p+1=0
p =-1
sin x = -1
xo = 270o
2
90o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Higher
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Harder Example: Solve
Step 1: Re-Arrange
3sin 2 x  4sin x  1  0 (0  x  360o )
Step 2: Consider what solutions
are expected
(3sin x  1)(sin x  1)  0
1
sin x 
3

2
90o
sin x  1

Two solutions One solution
180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Higher
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Step 3: Solve the equation
1
sin x 
3
1st Q
x = 19.5o
2nd Q
x = 160.5o
Overall solution
sin x  1
x = 90o
x = 19.5o , 90o and 160.5o
Solving Trigonometric Equations
Higher
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Graphical solution for
3sin 2 x  4sin x  1  0
cos(2x – 10o) – 0.5 = 0
Range : 0 ≤ x ≤ 2π
tanxo – 0.5 = 0
Range : 0 ≤ x ≤ 180o
Name :
2sin2xo + 3sinxo +1 = 0
Higher Trig.
6cos2xo
Range : 0 ≤ x ≤ 2π
+3=0
Range : 0 ≤ x ≤ 360o
Range : 0 ≤ x ≤ 2π
sin2xo – 0.5 = 0
cos(2x – 10o) – 0.5 = 0
Range : 0 ≤ x ≤ 2π
cos(2xo – 10o) = 0.5
2xo
– 10 = cos
o
-1
S
T
(0.5)
2x o -10o = 60o ,300o , 420o , 660o
A
C
2x o = 70o ,310o , 430o , 670o
x o = 35o ,155o , 215o , 335o
xo =
7 31 43 67
,
,
,
36 36 36 36
tanxo – 0.5 = 0
Range : 0 ≤ x ≤ 180o
tanxo
= 0.5
xo = tan
xo
=
-1
26.5o
,
206.5o
S
T
Name :
A
C
-1
(0.5)
2π 4π 8π 10π
, , ,
3 3 3 3
π 2π 4π 5π
xo = , , ,
3 3 3 3
S A
T C
(2sinxo + 1 ) = 0
sinxo = -0.5
(sinxo + 1 ) = 0
sinxo = -1
xo = 210o , 330o
xo = 270o
6cos2xo + 3 = 0 Range : 0 ≤ x ≤ 2π
2xo = cos
Range : 0 ≤ x ≤ 360o
Let p = sinxo
2p2 + 3p + 1 = 0
(2p + 1 )(p + 1) = 0
Higher Trig.
cos2xo = -0.5
(0.5)
2sin2xo + 3sinxo + 1 = 0
1st
Quad
xo =
sin2xo – 0.5 = 0
sin
= 0.5

sinxo
3
sinxo = 1 /√2
2x o =
xo =
S
T
xo
2
A
C
= ±1/√2
π 3π
,
4 4
Range : 0 ≤ x ≤ 2π
1st Quad
xo
=

4
sinxo = - 1/√2
xo =
S A
T C
5π 7π
,
4 4
Trigonometric Equations
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Higher
Double angle formulae (like cos2A or sin2A) often occur in
trig equations. We can solve these equations by substituting
the expressions derived in the previous sections.
Rules for solving equations
sin2A = 2sinAcosA when replacing sin2Aequation
cos2A = 2cos2A – 1
if cosA is also in the equation
cos2A = 1 – 2sin2A if sinA is also in the equation
Trigonometric Equations
Higher
cos 2 x o  4sin x o  5  0 for 0  x  360o.
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Solve:
(1  2sin 2 x)  4sin x  5  0
6  4sin x  2sin x  0
2
cos2x and sin x,
so substitute 1-2sin2x
compare with 6  4 z  2 z 2  0
(6  2sin x)(1  sin x)  0
sin x  1 or sin x  3
x  90
o
0  sin x  1 for all real angles
Trigonometric Equations
Higher
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Solve:
5cos 2 x o  cos x o  2
for 0  x  360o
cos 2x and cos x,
so substitute 2cos2 -1
5(2cos 2 x  1)  cos x  2

2
10cos 2 x  cos x  3  0
(5cos x  3)(2cos x  1)  0
3
1
cos x 
or cos x  
5
2

x  180  60  120
x  53.1o and
x  360  53.1  306.9
90o
o
o
x  180  60  240o
o
180
and
S
A
T
C
270o
3
2
0o
Trigonometric Equations
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Higher
The solution is to be in radians – but
work in degrees and convert at the end.
Solving Trigonometric Equations
Higher
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Harder Example:
Solve 5sin 2 x  2  2cos x
Step 1: Re-Arrange
(0  x  2 )
Step 2: Consider what solutions
are expected
5(1  cos 2 x)  2  2cos x

2
Remember othis !
90
2
2
3  2cos x  5cos 2 x  0
(3  5cos x)(1  cos x)  0
3
cos x 
5
cos x  1
sin   cos   1

2
cos 2   1  sin

A
S
o 2
180sin
  1  cos 2 
C
T
270o
Two solutions One solution
3
2
0o
Solving Trigonometric Equations
Higher
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Step 3: Solve the equation
3
cos x 
5
1st Q
4th Q
x = 53.1o
x = 306.9o
Overall solution in radians
cos x  1
x = 180o
x = 0.93 , π and 5.35
Solving Trigonometric Equations
Higher
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Graphical solution for
5 sin 2 x  2  2cos x
Maths4Scotland
Higher
Solve the equation 3cos(2 x)  10cos( x)  1  0 for 0 ≤ x ≤  correct to 2 decimal places
Replace cos 2x with
Substitute
Simplify
cos 2 x  2 cos 2 x  1
3  2 cos x  1  10 cos x  1  0
2
6 cos x  10 cos x  4  0
2
Determine quadrants
S
A
T
C
3cos 2 x  5cos x  2  0
Factorise
Hence
3cos x 1 cos x  2  0
cos x 
x  1.23
1
3
cos x  2 Discard
Find acute x
Previous
acute
x  1.23
x  5.05
x  1.23 rad
Quit
or
2  1.23
rads
rads
rads
Hint
Quit
Next
Maths4Scotland
Higher
Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x
a) Find expressions for:
i) f(g(x))
ii) g(f(x))
Determine x
b) Solve 2 f(g(x)) = g(f(x)) for 0  x  360°
sin x  0  x  0, 360
1st expression
f ( g ( x))  f (2 x)  sin 2 x
2nd expression
g ( f ( x))  g (sin x)  2sin x
Form equation
Replace sin 2x
2sin 2x  2sin x  sin 2 x  sin x
Common factor
1
2

acute
x  60
S
A
T
C
Determine
quadrants
x  60, 300
2sin x cos x  sin x
x  0, 60, 300, 360
2sin x cos x  sin x  0
Rearrange
Hence
cos x 
sin x  2cos x 1  0
sin x  0
or
2 cos x  1  0  cos x 
Previous
Table of exact values
Quit
Hint
1
2
Quit
Next
Maths4Scotland
Higher
Functions f ( x)  sin x, g ( x)  cos x
a)
Find expressions for
b)
i)
2nd
i) f(h(x))

4
are defined on a suitable set of real numbers
ii) g(h(x))
1
1
f (h( x)) 
sin x 
cos x
2
Hence solve the equation2
expression
expression
Simplify
h( x )  x 
Show that
iii)
1st
and
1st
ii) Find a similar expression for g(h(x))
f (h( x))  g (h( x))  1 for 0  x  2

 sin x 
Simplifies to
4
   
g (h( x))  g  x    cos  x  

f (h( x))  f x 
4


4
4


4
4
f (h( x))  sin x cos  cos x sin
expr.
Rearrange:
1
1
sin x 
2
2
Use exact values
f (h( x)) 
Similarly for 2nd expr.
g (h( x))  cos x cos  sin x sin
g (h( x)) 
Form Eqn.
1
2
acute x
cos x


4
4
cos x 
1
sin x
2
f (h( x))  g (h( x))  1
Previous
Table of exact values
Quit
Quit
acute
Determine
quadrants
x
 3
4
,
sin x 
2
2
sin x  1
2

2
x
2
1

2 2
2

4
S
A
T
C
4
Hint
Next
Maths4Scotland
a)
b)
Higher
Solve the equation sin 2x - cos x = 0 in the interval 0  x  180°
The diagram shows parts of two trigonometric graphs,
y = sin 2x and y = cos x. Use your solutions in (a) to
write down the co-ordinates of the point P.
Replace sin 2x
Common factor
Hence
2sin x cos x  cos x  0
cos x  2sin x 1  0
cos x  0
Determine x
or
Solutions for where graphs cross
x  30, 90, 150
2sin x  1  0  sin x 
1
2
cos x  0  x  90, ( 270 out of range)
sin x 
1
2

acute
x  30
S
A
Determine quadrants
for sin x
Previous
Table of exact values
x  150
y  cos150
Find y value
y
Coords, P
x  30, 150
T
By inspection (P)

P 150, 
3
2

Hint
C
Quit
Quit
3
2
Next
Maths4Scotland
Solve the equation
Higher
3cos(2 x)  cos( x)  1
for 0 ≤ x ≤ 360°
cos 2 x  2 cos 2 x  1
Replace cos 2x with
Determine quadrants
3  2 cos x  1  cos x  1
2
Substitute
Simplify
6 cos 2 x  cos x  2  0
Factorise
3cos x  2 2cos x 1  0
cos x  
Hence
Find acute x
acute
2
3
x  48
cos x 
acute
1
2
x  60
cos x  
2
3
cos x 
acute
x  48
acute
x  60
S
A
S
A
T
C
x  132
x  228
T
C
x  60
x  300
Solutions are: x= 60°, 132°, 228° and 300°
Previous
Table of exact values
Quit
Quit
1
2
Hint
Next
Maths4Scotland
Higher



Solve the equation 2sin 2 x  6  1
Rearrange
sin
Find acute x
Note range


2x 
6
acute

2x 


6
for 0 ≤ x ≤ 2




Determine quadrants
2x 

6


and for range
6

0  x  2  0  2 x  4
S
0  2 x  2
for range
1

2
2x 

6




6
2x 

6


5
6


17
6
2  2 x  4

13
6
2x 

6
A
Solutions are:
T
x
C

6
,

2
,
7 3
,
6
2
Hint
Previous
Table of exact values
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Quit
Next
Maths4Scotland
Higher
a) Write the equation cos 2q + 8 cos q + 9 = 0 in terms of cos q
and show that for cos q it has equal roots.
b) Show that there are no real roots for q
Replace cos 2q with
cos 2q  2 cos 2 q  1
Rearrange
2 cos 2 q  8cos q  8  0
Divide by 2
cos 2 q  4 cos q  4  0
Factorise
Deduction
Try to solve:
 cosq  2  0
cosq  2
No solution
Hence there are no real solutions for q
 cosq  2 cosq  2  0
Equal roots for cos q
Hint
Previous
Quit
Quit
Next
Maths4Scotland
Higher
Solve algebraically, the equation sin 2x + sin x = 0, 0  x  360
Replace sin 2x
2sin x cos x  sin x  0
Common factor
sin x  2cos x  1  0
Hence
S
A
T
C
sin x  0
or
Determine x
Determine quadrants
for cos x
2 cos x  1  0

cos x  
1
2
x  120, 240
sin x  0  x  0, 360
1
2
cos x   
acute
x  60
x = 0°,
Previous
Table of exact values
Quit
120°, 240°, 360°
Quit
Hint
Next
Maths4Scotland
Higher
Solve the equation
cos2x  cos x  0
Replace cos 2x with
cos 2 x  2 cos 2 x  1
for 0 ≤ x ≤ 360°
Determine quadrants
cos x 
2 cos 2 x  1  cos x  0
Substitute
Simplify
2 cos 2 x  cos x  1  0
Factorise
 2cos x 1 cos x  1  0
cos x 
Hence
Find acute x
acute
1
2
x  60
cos x  1
1
2
acute
x  60
S
A
T
C
x  60
x  300
x  180
Solutions are: x= 60°, 180° and 300°
Previous
Table of exact values
Quit
Quit
Hint
Next
Maths4Scotland
Higher
cos2x  5cos x  2  0
Solve algebraically, the equation
Replace cos 2x with
Substitute
cos 2 x  2 cos 2 x  1
for 0 ≤ x ≤ 360°
Determine quadrants
2 cos x  1  5cos x  2  0
2
cos x 
acute
Simplify
2 cos 2 x  5cos x  3  0
Factorise
 2cos x 1 cos x  3  0
Hence
Find acute x
cos x 
acute
1
2
x  60
S
cos x  3
Discard above
1
2
x  60
A
T
C
x  60
x  300
Solutions are: x= 60° and 300°
Previous
Table of exact values
Quit
Quit
Hint
Next
Maths4Scotland
Higher
f ( x)  2 cos x  3sin x
a)
Express f (x) in the form k cos( x   )
b)
Hence solve algebraically f ( x)  0.5
Expand kcos(x - a):
k cos a  2
Square and add
k 2  22  32
Put together:
tan a 
 x  56  82
Previous
and
0    360
0  x  360
3
2
acute
k sin a  3
k  13
a  56
a is in 1st quadrant
(sin and cos are both + )
a  56
2cos x  3sin x  13 cos  x  56 
Solve equation.
acute
for
k 0
k cos( x  a)  k cos x cos a  k sin x sin a
Equate coefficients:
Dividing:
where
cos  x  56  
13 cos  x  56   0.5
Cosine +, so 1st & 4th quadrants
Quit
Quit
0.5
13
x  138 or x  334
Next
Hint
Maths4Scotland
Higher
Solve the equation
2sin x  3cos x  2.5 in the interval 0  x  360.
Use R cos(x - a):
R cos( x  a)  R cos x cos a  R sin x sin a
R cos a  3
Equate coefficients:
R 2  22   3
Square and add
tan a  
Dividing:
Put together:
 x 146  46
x  192
or
Previous
acute
R  13
a  34
a is in 2nd quadrant
a  146
(sin + and cos - )
2sin x  3cos x  13 cos  x  146 
Solve equation.
acute
2
3
2
R sin a  2
x  460
cos  x  146 
13 cos  x  146  2.5
2.5
13
Cosine +, so 1st & 4th quadrants
(out of range, so subtract 360°)
Quit
x  100
Quit
or
x  192
Hint
Next
Maths4Scotland
Higher
Solve the simultaneous equations
k sin x  5
k cos x  2
where k > 0 and 0  x  360
Use tan A = sin A / cos A
Divide
tan x 
Find acute angle
5
2
acute
Determine quadrant(s)
x  68
Sine and cosine are both + in original equations
Solution must be in 1st quadrant
State solution
x  68
Hint
Previous
Quit
Quit
Next
Solving Trig
Higher
Example
Expand and
Equations equate
coefficients
Outcome 4
www.mathsrevision.com
a) Express 2cos 2 x  3sin 2 x in the form k sin(2 x   )
b) Hence solve 2cos 2 x  3sin 2 x  1 for 0  x  360o

Find tan ratio
note:2
o
sin(-) and90
cos(-)
 3sin 2 x  2cos 2 x  k sin(2 x   )
 k sin 2 x cos  k cos2 x sin 
k sin   2 

k cos   3
Square and add
k  13
2
sin 
2
 tan   tan 1    33.7o
cos 
3
  180o  33.7o  213.7o

180
o
S
A
T
C
270o
3
2
0o
Solving Trig Equations
Outcome 4
www.mathsrevision.com
Higher
b) We now have 2cos 2 x  3sin 2 x  13 sin(2 x  213.7)0
We solve
2cos 2 x  3sin 2 x  1
by solving
13 sin(2 x  213.7)  1
sin(2 x  213.7) 
1
13
In the 1st quadrant
2x – 213.7 = 16.1o
163.9o
376.1o
523.9o
x = 114.9o ,
188.8o,
 1 
0
sin 1 

16.1

 13 
o
o
2x = 229.8
377.6
368.8º - 360º
589.8o= 8.8º
737.6o
294.9o,
368.8o
Range : 0 ≤ x ≤ π
Range : 0 ≤ x ≤ 360o
Solving Trig. Equations
Range : 0 ≤ x ≤ 2π
Name :
Are you on Target !
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Higher
•
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•
Make sure you complete and correct
MOST of the Trigonometry
questions in the past paper booklet.