1. Trigonometric Equations
Lec f 12
Trigonometry
2.4 Solutions Of
Trigonometric Equations

2. Learning Outcomes: to solve equation using half-angle identity
for sin θ, cos and tan in terms of
Express a cos + b sin as R cos ( )
or Rsin ( ) and subsequently solve
the equation a cos θ + b sin θ = c
Determine the maximum and minimum
value of trigonometric expressions.

3. The Substitution
This method is very useful in solving
trigonometric equations which cannot be
solved by using identities.
Remember the half-angle formula:

4. By using Phytogaras Theorem:
OP= 1+ t2 2t
1- t2 P O Q q
1+ t2

5. By using these substitutions, the
trigonometric equation is reduced to an
algebraic equation in terms of t.
It can be used if each term in the
equation has the same angle.

6. Example 1
Solve the equation 5 tan + sec + 5 = 0 for
using

7. 5 tan + sec = 0
Solution:
Same angle

8. 71.560
θ/2
26.560

9. Example 2
Solve the equation 5 cos - 2sin = 2 for

10. Solution:
5 cos - 2sin = 2
Same angle

11. 0.4049
0.7854
θ/2

12. Example 3
Solve the equation 2 sin cos2 = -1 for

13. 2 sin2 - cos2 = -1
Solution:
Same angle

14. y x
63.43

15. Equation in the form acos + bsin = c
This type of equation can be solved by using 2 different methods:
Using the substitution
( as the previous examples)
(ii) Express as Rcos ( ) or
Rsin( ) and subsequently
solve the equation

16. NOTES:

17. NOTES:

18. Example 4
By expressing 2 sin θ + 5 cos θ in the form Rsin ( θ + α ), hence solve the equation 2 sin θ + 5 cos θ = - 3 for
00 < θ < 3600

19. Solution:
Let 2 sin θ + 5 cos θ ≡ R sin ( θ + α ); R > 0
2 sin θ + 5 cos θ ≡ R [sin θ cos α + cos θsin α ]
2 sin θ + 5 cos θ ≡ R sin θ cos α + R cos θsin α
Equating the coefficient of sin θ and cos θ :
R sin α = 5
1st quadrant
R cos α = 2

20. 5 2 R α
α = 68.20
2 sin θ + 5 cos θ = sin (θ );

21. 2sinθ + 5 cosθ = - 3; 00 <θ < 3600
33.9
θ = 213.9, 326.1
θ = ,

22. Example 5
By expressing cos 2x – sin 2x in the
form R sin ( 2x + α ) where R > 0 and
0 < α < 2π, solve the equation
cos 2x – sin 2x = 1 for

23. cos 2x - sin 2x ≡ R sin ( 2x + α ); R > 0
Solution:
cos 2x - sin 2x ≡ R sin ( 2x + α ); R > 0
cos 2x - sin 2x ≡ R [sin 2x cos α + cos2x sin α]
cos 2x - sin 2x ≡ Rsin 2x cos α + Rcos2x sin α
Equating the coefficient of sin 2x and cos 2x :
(i)
(ii)
R sin α =
2nd quadrant
R cos α = -1

24. R 1
cos 2x - sin 2x = 2 sin ( 2x )

25. cos 2x - sin 2x = 1
2 sin ( 2x ) = 1
sin ( 2x ) =

27. Maximum and Minimum Values
of Trigonometric Expressions

28. Example 6
Find the maximum and minimum values
( the greatest and the least values ) for:
2 sin θ + 5 cos θ
(b) cos 2x – sin 2x

29. Solution:
(a) 2 sin θ + 5 cos θ = sin (θ ) [ Eg. 4 ]
Max = 1
Min = -1
Maximum value = (1) =
Minimum value = (-1) = -

30. (b)
cos 2x - sin 2x = 2 sin ( 2x ) [ Eg. 5]
Max = 1
Min = -1
Maximum value = 2 (1) = 2
Minimum value = 2 (-1) = - 2

31. Example 7
Find the maximum and minimum value for
cos 2x + sin 2x. Hence, find the value
of x when that expression is maximum and
minimum for

32. Solution:
cos 2x + sin 2x =
2 – ( cos 2x- sin2x)
[ Eg 5 ]
2 sin ( 2x )
= 2 –
Max = 1
Min = -1
Maximum value = 4
Minimum value = 0
Max = - 2
Min = 2

34. When the expression is maximum:
2 sin ( 2x ) = 4
2 –
Thus the value of x when the expression is maximum are:
sin ( 2x ) = - 1

35. 2 sin ( 2x + ) = 0 2 – 2 sin ( 2x + ) = 0 2 – sin ( 2x + ) = 1
When the expression is minimum:
2 sin ( 2x ) = 0
2 –
2 sin ( 2x ) = 0
2 –
Thus the value of x when the expression is minimum are:
sin ( 2x ) = 1

36. Excercise
By expressing 5 cos x + 3 sin x in the
form R cos ( x - α ), find the maximum
value for 5 cos x + 3 sin x. Solve the
equation 10 cos x + 6 sin x = 7 for values
of x lying between 0 and 2л. Give your
answer in two decimal places.

37. Solution:
Let 5 cos x + 3 sin x ≡ R cos ( x - α );R > 0
5 cos x + 3 sin x ≡ R cosx cos α + Rsinx sin α
Equating the coefficient of cos x and sin x :
R cos α = 5
1st quadrant
R sin α = 3

38. R 3 5 α R 3 5 α R 3 5 α

39. 5 cos x + 3 sin x = cos ( x – )
Maximum value for
5 cos x + 3 sin x = (1) =
0 < x < 2л
10 cos x + 6 sin x = 7;
5 cos x + 3 sin x = 3.5

40. cos ( x – ) = 3.5
cos ( x – ) =
cos ( x – ) =
0.9270
x – = ,
x = 1.467,
x = 1.47, 5.90 ( 2 dp)

41. Conclusion
Trigonometric Equation in the form
a cos θ + b sin θ = c can be solved by using:
Substitution
Compound angle formulae
(b) To find the minimum and maximum
value for trigonometris expression,
express acos θ+bsinθ as Rcos ( ) or Rsin( )