1. Solving
Trig
Equations

2. Graph of y = sin x
y = sin x
Max sin x = 1 when x = 90°
Min sin x = -1 when x = 270°

3. Graph of y = cos x
y = cos x
Max cos x = 1 when x = 0° or 360°
Min cos x = -1 when x = 180°

4. Graph of y = tan x
These lines are called asymptotes and show that tan xo is undefined for 90o and 270o
The graph of y = tan xo has no maximum or minimum values

5. A S T C +ve +ve +ve +ve –ve –ve –ve –ve
Splitting each graph into 90o sections, shows that each has two positive and two negative sections
90o
+ve
+ve A S
sin
all
180o 0o
360o
–ve
–ve T C
tan
cos
All Sinners Take Care
270o

6. The diagram shows where the various ratios are positive
90°
90° to 180° sin positive
0° to 90° all positive S A
180 -
180°
0°,360°
180 + C
360 - T
180° to 270° tan positive
270° to 360° cos positive
270°

7. Solve the following equations, giving two values
for x between 0° and 360°.
cos x° = 0·085 tan x° = 0·510
+ve
-ve
Ignore sign
90o A  S  A S 0o
180o 0o
180o
360o 
360o T C  T C
270o
270o

8. Trigonometric Equations Solve the following equations,
giving two values
for x between 0° and 360°.
sin x° = 0·766 cos x° = 0·565
tan x° = 4·915 cos x° = 0·906
sin x° = 0·707 tan x° = 2·050
sin x° = 0·415 tan x° = 0·193

9. Trigonometric Equations Solutions
sin x° = 0·766
x = 50·0 or 130·0 cos x° = 0·565
x = 55·6 or 304·4
tan x° = 4·915
x = 78·5 or 258·5 cos x° = 0·906
x = 155·0 or 205·0
sin x° = 0·707
x = 225·0 or 315·0
tan x° = 2·050
x = 116·0 or 296·0
sin x° = 0·415
x = 24·5 or 155·5
tan x° = 0·193
x = 10·9 or 190·9

10. Solve the following equations, giving two values
for x between 0° and 360°.
3sin x°  1 = tan x° + 5 = 1
2tan xo = – 4
3sin xo = 1
tan xo = – 2
sin xo = 1 ÷ 3 = 0∙333
tan-1 2
= 63∙4o A 0o
180o
270o
360o S T C
90o
sin-1 0∙333
= 19∙5o   
x = 180 – 63∙4o
x = 19∙5o
x = 116∙6o 
x = 180 – 19∙5o
x = 360 – 63∙4o
x = 160∙5o
x = 296∙6o

11. Trigonometric Equations
Solve the following equations, giving two values
for x between 0° and 360°.
4cos x°  2 = tan x°  12 = 0
5sin x° + 4 = cos x° + 3 = 0
3tan x° + 2 = tan x° - 5 = 0

12. 4cos x°  2 = 0
cos x° = 2  4 = 0·5
x = 60 or 300
5sin x° + 4 = 0
sin x° = 4  5 = 0·8
x = 233·1 or 306·9
3tan x° + 2 = 1
tan x° = 1  3 = 0·333.
x = 161·6 or 341·6
5tan x°  12 = 0
tan x° = 12  5 = 2·4
x = 67·4 or 247·4
4cos x° + 3 = 0
cos x° = 3  4 = 0·75
x = 138·6 or 221·4
2tan x° - 5 = 0
tan xo = 5 ÷ 2 = 2∙5
x = 68∙2 or 248∙2

13.   The diagram shows part of the graph of y = 4cos x° and y = – 3
Find the coordinates of the points A and B A B
(138·6, -3)
(221·4, -3) A 0o
180o
270o
360o S T C
90o 
Where the graphs intersect y = y
 4cos xo = – 3 
 cos xo = – 3/4
 x = ·4o
 cos-1 ¾ = 41·4o
 x = ·4o

14. The maximum value of the sine of any angle is 1 and the minimum – 1
Evaluating Trig Functions
The height, h km, of a satellite above the earth after x hours is given by : h(x) = sin (10x)o
a) What are the maximum and minimum heights of the satellite?
b) When do these maximum and minimum heights occur?
c) What is the period of h?
d) Draw a sketch of h(x) for 0 ≤ x ≤ 180
The maximum value of the sine of any angle is 1 and the minimum – 1
Max h(x) = × 1
= 840 km
Min h(x) = × (– 1)
= 760 km

15. Since all values repeat every 360o
Max h(x) = × 1
= 840 km
Occurs when angle = 90o
Min h(x) = × (– 1)
= 760 km
Note: When 0 ≤ x ≤ 180, ≤ 10x ≤ 1800
Max h(x) occurs when 10x = 90, 450, 810, 1170, ……..
Since all values repeat every 360o
Max h(x) occurs when x = 9 hrs, 45 hrs, 81 hrs, 117 hrs, ……..