Trigonometric Equations : Session 1

Illustrative Problem Solve :sinx + cosx = 2 Solution:

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

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 

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

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

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

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

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

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

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

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.

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.

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,

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.

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,

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

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,

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

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

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

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

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

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

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

Illustrative problem Find the solution of sinx = Solution:

Illustrative problem Solve tan2x = Solution: We have tan2x =

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

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

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

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

Illustrative Problem Solve :sinx + siny = 2 Solution:

Class Exercise Q1. Solve :sin5x = cos2x Solution:

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

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

Class Exercise Q4. Solve : Solution:

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

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

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

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

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

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

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

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:

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

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

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