1. TRIGONOMETRIC EQUATIONS
TEACHER NOTE:
Only degrees are used in these slides. Questions are also asked on radians, so teachers should make sure to show how to solve questions using radians in other examples.

2. Firstly recall the graphs of y = sin x, y = cos x and y = tan x.
Note: sin x is positive
for 0º < x < 180º,
and negative for
180º < x < 360º.
y = sin x
Note: cos x is positive
for 0º ≤ x < 90º, and
for 270º < x ≤ 360º
and negative for
90º < x < 270º.
y = cos x
y = tan x
Note: tan x is positive
for 0° < x < 90º, and
for 180º < x < 270º
and negative for
90º < x < 180º. and
for 270º < x < 360º.

3. S A T C We can summarise the information in the following diagram:
0º, 360º
180º
90º
270º
sin +ve
cos +ve
tan +ve
sin +ve
cos –ve
tan –ve
sin –ve
cos –ve
tan +ve
sin –ve
cos +ve
This can be simplified to show just the positive ratios:
0º, 360º
180º
90º
270º
all +ve
sin +ve
tan +ve
cos +ve A S T C
TEACHER NOTE:
CAST diagram used here – teacher could also use sin, cos and tan graphs here.
or just:

4. S A T C The positive ratio diagram can be used to solve
trigonometric equations:
Example 1:
Solve cos x = 0.5; 0º ≤ x < 360º A S T C
We can see that the cosine of an
angle is positive in the two quadrants
on the right, i.e. the 1st and 4th
quadrants.
Using: 
α = cos–1 0.5
= 60º
Hence the solutions in the given range are:
x =
60°,
360 – 60°
so x = 60º, 300º

5. Example 2: Solve sin x = –0.5; –360º < x ≤ 360º
The sine of an angle is negative in the
3rd and 4th quadrants. 
α = sin–1 0.5
= 30º
N.B. The negative is ignored
to find the acute angle.
Hence the solutions in the given range are:
x =
–150°,
–30°,
210°,
330°
In problems where the angle is not simply x, the given range
will need to be adjusted.

6. Example 3: Solve 3 + 5 tan 2x = 0; 0º ≤ x ≤ 360º.
Firstly we need to make tan 2x the subject of the equation: – 3 5
tan 2x =
The tangent of an angle is negative in the
2nd and 4th quadrants. 
= 30.96°
The range must be adjusted for the angle 2x.
i.e. 0° ≤ 2x ≤ 720°.
Hence: 2x =
149.04°,
329.04°,
509.04°,
689.04°.
x = 74.5°, 164.5°, °, 344.5°.

7. Example 4: Solve 2 sin (4x + 90º) – 1 = 0; 0 < x < 90º
The sine of an angle is positive in the
1st and 2nd quadrants. 
= 30º
The range must be adjusted for the angle 4x + 90º.
i.e. 0º < 4x < 360º
90º < 4x + 90º < 450º
4x + 90º =
150°,
390°
4x = 60º, 300º
x = 15º, 75º

8. Example 5: Solve 6 sin2 x + sin x – 1 = 0; 0º ≤ x < 360º
A quadratic equation!
It may help to abbreviate sin x with s:
i.e. 6s2 + s – 1 = 0
Factorising this:
(3s – 1)(2s + 1)= 0  
α = 19.47º
α = 30º
So, x =
19.5°,
160.5°,
210°,
330°.
(To nearest 0.1º.)

9. A S T C Summary of key points: To solve a Trigonometric Equation:
Re-arrange the equation to make sin, cos or tan of
some angle the subject.
Locate the quadrants in which the ratio is positive,
or negative as required. A S T C
Using:
Adjust the range for the given angle.
Read off all the solutions within the range.