2. Trigonometric Equations : Session 1

3. Illustrative Problem
Solve :sinx + cosx = 2
Solution:

4. Definition A trigonometric equation is an equation
Contains trigonometric functions of variable angle
sin  = ½
2 sin2 + sin22 = 2.

5. Periodicity and general solution
Solution of Trigonometric Equation:
Values of , which satisfy the trigonometric equation
For sin  = ½ ,  =  /6, 5  /6, 13  /6,…….
No. of solutions are infinite .
Why ? - Periodicity of trigonometric functions.
e.g. - sin, cos  have a period as 2 

6. Periodicity and general solution
Periodicity of trigonometric functions.
f(+T) = f() 
sin  2 3 4
sin have a period 2 

7. Graph of y=sinx sinx is periodic of period 2  -2  2 - Y X (0,1)
-2  2 - Y X
(0,1)
(0,-1)

8. Graph of y=cosx cosx is periodic of period 2  -2  2 X Y (0,1)
(0,-1) Y

9. Graph of y=tanx tanx is periodic of period 
tanx is not defined at x=(2
where n is integer
n+1)
-2  2 X Y

10. Periodicity and general solution
As solutions are infinite , the entire set of solution can be written in a compact form.
This compact form is referred to as general solution
For sin  = ½ ,  = /6, 5/6, 13/6,…….
Or  = n  +(-1)n ( /6)
General Solution

11. Principal Solutions
Solutions in 0x2
 principal solutions.

12. Illustrative problem Find the principal solutions of tanx = Solution
We know that tan( - /6) =
and tan(2 - /6) =
 principal solutions are 5/6 and 11/6

13. Illustrative problem
Find the principal solution of the equation sinx = 1/2
Solution
sin/6 = 1/2
and sin(- /6) = 1/2
 principal solution are x = /6 and 5/6.

14. General solution of sin  = 0
sin = PM/OP
For sin = 0 , PM = 0 Y  X O P M X’ Y’
For PM = 0, OP will lie on XOX’
  = 0, ±π, ±2π, ±3π …..
  is an integral multiple of π.

15. General solution of sin  = 0
For sin  = 0 ,
 is an integer multiple of π.
Or  = nπ , n є Z (n belongs to set of integers)
Hence, general solution of sin = 0 is
 = nπ , where n є Z,

16. General solution of cos  = 0
cos  = OM/OP
For cos  = 0 , OM = 0 Y  X O P M X’ Y’
For OM = 0, OP will lie on YOY’
  = ±π/2, ±3π/2, ±5π/2….
 is an odd integer multiple of π/2.

17. General solution of cos  = 0
For cos  = 0 ,
 is an odd integer multiple of π/2.
Or  = (2n+1)π/2 , n є Z (n belongs to set of integers)
Hence, general solution of cos = 0 is
 = (2n+1)π/2 , where n є Z,

18. General solution of tan  = 0
tan = PM/OM
For tan = 0 , PM = 0 Y  X O P M X’ Y’
For PM = 0, OP will lie on XOX’
  = 0,±π, ±2π, ±3π….
 is an integer multiple of π.
Same as sin  = 0

19. General solution of tan  = 0
For tan  = 0 ,
 is an integer multiple of π.
Or  = nπ , n є Z (n belongs to set of integers)
Hence, general solution of tan = 0 is
 = nπ, where n є Z,

20. Illustrative Problem Find the general value of x
satisfying the equation sin5x = 0
Solution:
sin5x = 0 = sin0
=> 5x = n
=> x = n/5
=>x = n/5 where n is an integer

21. General solution of sin = k
If sin = k  -1 k  1
Let k = sin, choose value of  between –/2 to  /2
If sin = sin  sin - sin = 0

22. General solution of sin = k
Even , +ve
Odd , -ve
 = nπ +(-1)n , where n є Z

23. General solution of cos = k
If cos = k  -1 k  1
Let k = cos, choose value of  between 0 to 
If cos = cos  cos - cos = 0

24. General solution of cos = k
-ve
+ve
 = 2nπ   , where n є Z

25. General solution of tan = k
If tan = k  -  < k < 
Let k = tan, choose value of  between - /2 to /2
If tan = tan  tan - tan = 0
 sin.cos - cos.sin = 0

26. General solution of tan = k
sin.cos - cos.sin = 0
sin(  -  ) = 0
 -  = nπ , where n є Z
  = nπ + 
 = nπ+  , where n є Z

27. Illustrative problem
Find the solution of sinx =
Solution:

28. Illustrative problem
Solve tan2x =
Solution:
We have tan2x =

29. Illustrative problem
Solve sin2x + sin4x + sin6x = 0
Solution:

30. Illustrative Problem
Solve 2cos2x + 3sinx = 0
Solution:

31. General solution of sin2x = sin2 cos2x = cos2, tan2x = tan2
n   where n is an integer.

32. Illustrative Problem
Solve : 4cos3x-cosx = 0
Solution:

33. Illustrative Problem
Solve :sinx + siny = 2
Solution:

34. Class Exercise Q1.
Solve :sin5x = cos2x
Solution:

35. Class Exercise Q2.
Solve :2sinx + 3cosx=5
Solution:

36. Class Exercise Q3.
Solve :7cos2 +3sin2 = 4
Solution:

37. Class Exercise Q4.
Solve :
Solution:

38. Class Exercise Q5.
Show that 2cos2(x/2)sin2x = x2+x-2 for 0<x</2 has no real solution.
Solution:

39. Class Exercise Q6.
Find the value(s) of x in (- , ) which satisfy the following equation
Solution:

40. Class Exercise Q6. Solution:
Find the value(s) of x in (- , ) which satisfy the following equation
Solution:

41. Class Exercise Q7. Solve the equation sinx + cosx = 1+sinxcosx
Solution:

42. Class Exercise Q7. Solution: Solve the equation
sinx + cosx = 1+sinxcosx
Solution:

43. Class Exercise Q8.
If rsinx=3,r=4(1+sinx),
then x is
Solution:

44. Class Exercise Q8.
If rsinx=3,r=4(1+sinx),
then x is
Solution:

45. Class Exercise Q9.
In a  ABC ,  A >  B and if  A and  B satisfy
3 sin x –4 sin3x – k = 0 ( 0< |k| < 1 ) ,  C is
Solution:

46. Class Exercise Q10.
Solve the equation
(1-tan)(1+sin2) = 1+tan
Solution:

47. Class Exercise Q10. Solution: Solve the equation
(1-tan)(1+sin2) = 1+tan
Solution:

48. Class Exercise Q10. Solution: Solve the equation
(1-tan)(1+sin2) = 1+tan
Solution: