Trigonometry
Solution Of Trig Equations

Cases Where We Only Have
sin, cos Or tan In An Equation
And
sin, cos Or tan
Is Already
Equal To A Number

sinθ = a positive value

Using the quadrant diagram to solve cases where
sinθ = a positive value

Example
sinθ = 0.5

sinθ = a negative value

Using the quadrant diagram to solve cases where
sinθ = a negative value

Example
sinθ = - 0.5

cosθ = a positive value

Using the quadrant diagram to solve cases where
cosθ = a positive value

Example
cosθ = 0.5

cosθ = a negative value

Using the quadrant diagram to solve cases where
cosθ = a negative value

Example
cosθ = - 0.5

tanθ = a positive value

Using the quadrant diagram to solve cases where
tanθ = a positive value

Example
tanθ = √3

tanθ = a negative value

Using the quadrant diagram to solve cases where
tanθ = a negative value

Example
tanθ = - √3

Cases Where We Only Have
sin, cos Or tan In An Equation
But
Manipulation Is Required
To
Obtain sin, cos Or tan
Equal To A Number

[1] 2cosθ - √3 = 0
[2] 4sint - √3 = 2sint
[3] 2cosθ - 5 = 3cosθ - 2

Cases Where We Have
Both sin And cos In An Equation
But
We Can Manipulate
To Obtain
tan Equal To A Number

Example
2sinθ = cosθ
This can be manipulated to give
tanθ = 0.5

Example
4sinθ - cosθ = 0
This can be manipulated to give
tanθ = 0.25

Example
5sinθ + 2cosθ = 0
This can be manipulated to give
tanθ = - 0.4

Cases Where We Have
Two Out Of Three Of
sin Or cos Or tan In An Equation
But
Where We Can Use
Highest Common Factor Factorisation
To
Obtain sin And/Or cos And/Or tan
Equal To A Number

Example
0.5sinθ + sinθcosθ = 0
sinθ(0.5 + cosθ) = 0
sinθ = 0
(0.5 + cosθ) = 0 => cosθ = - 0.5

Example
sinθ + sinθcosθ = 0
sinθ(1 + cosθ) = 0
sinθ = 0
(1 + cosθ) = 0 => cosθ = - 1

Example
0.2sinθ = sinθcosθ
0.2sinθ - sinθcosθ = 0
sinθ(0.2 - cosθ) = 0
sinθ = 0
(0.2 - cosθ) = 0 => cosθ = 0.2

Cases Where We Have
A Quadratic Equation
Involving sin Or cos Or tan
But
We Have No Number Term
Therefore
We Can Use
Highest Common Factor Factorisation
To
Obtain sin Or cos Or tan
Equal To A Number

Example
sin²θ + 0.5sinθ = 0
sinθ(sinθ + 0.5) = 0
sinθ = 0
(sinθ + 0.5) = 0 => sinθ = - 0.5

Example
cos²θ = 0.3cosθ
cos²θ - 0.3cosθ = 0
cosθ(cosθ - 0.3) = 0
cosθ = 0
(cosθ - 0.3) = 0 => cosθ = 0.3

Example
sin²θ + sinθ = 0
sinθ(sinθ + 1) = 0
sinθ = 0
(sinθ + 1) = 0 => sinθ = - 1

Cases Where We Have
A Quadratic Equation
Involving sin Or cos Or tan
And
A Number Term Is Included In The Equation
Therefore
We Need To Use
Quadratic Factorisation
To
Obtain sin Or cos Or tan
Equal To A Number

Example
sin²θ + 3sinθ + 2= 0
(sinθ + 2)(sinθ + 1) = 0
(sinθ + 2) = 0 => sinθ = - 2 => NO real values of θ
(sinθ + 1) = 0 => sinθ = - 1 => θ = 270° => θ = 270n° (n = integer)

Examples On Video

[1] Highest Common Factor Factorisation
sinx + 2sinxcosx = 0

[2] Quadratic Factorisation
2cos²θ + 11cosθ = - 11

Cases Where
The Variable To Be Solved For
Has Been Transform Before
The sin Or cos Or tan Function Has been Applied

METHOD
We need to 'solve for the bracket'
and then adjust

Example
sin(θ + 20) = 0.5
(θ + 20) = invsin(0.5)
(θ + 20) = 30°, 150°, 390°, 510°, etc
θ = (30 - 20)°, (150 - 20)°, (390 - 20)°, (510 - 20)°, etc
θ = 10°, 130°, 370°, 490°, etc

Example
sin(2θ + 20) = 0.5
(2θ + 20) = invsin(0.5)
(2θ + 20) = 30°, 150°, 390°, 510°, etc
2θ = (30 - 20)°, (150 - 20)°, (390 - 20)°, (510 - 20)°, etc
2θ = 10°, 130°, 370°, 490°, etc
θ = 5°, 65°, 185°, 245°, etc

Examples On Video

[1] Quadratic Factorisation
2sin²θ - 2sinθ - 1 = 0

[2] 'Solve For The Bracket' And Then Adjust
cos(A - 50) = (√3)/2

Example On Video

2sinxcosx = 0

Different Methods Shown

Examples On Video

[1] sinx + 2sinxcosx = 0

[2] sin²x - 1 = 0

A More Difficult Example On Video

METHOD
Factorisation By Grouping

2sin(t)cos(t) - cos(t) + 2sin(t) - 1 = 0