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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 + sin 2 2 = 2. Periodicity and general solution.
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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
sin 0 2 3 4 Periodicity and general solution Periodicity of trigonometric functions. f(+T) = f() sin have a period 2
sinx is periodic of period 2 Y (0,1) 0 X 2 -2 - (0,-1) Graph of y=sinx
cosx is periodic of period 2 Y (0,1) 0 X 2 -2 (0,-1) Graph of y=cosx
tanx is periodic of period n+1) tanx is not defined at x=(2 where n is integer Y 0 X 2 -2 Graph of y=tanx
Or = n +(-1)n ( /6) General Solution Periodicity and general solution As solutions areinfinite , 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,…….
Principal Solutions Solutions in 0x2 principal solutions.
We know that tan( - /6) = and tan(2 - /6) = Illustrative problem Find the principal solutions of tanx = Solution 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.
Y P X X’ O M Y’ General solution of sin = 0 sin = PM/OP For sin = 0 , PM = 0 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,
Y P X X’ O M Y’ General solution of cos = 0 cos = OM/OP For cos = 0 , OM = 0 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,
Y P X X’ O M Y’ General solution of tan = 0 tan = PM/OM For tan = 0 , PM = 0 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 = 2nπ + = (2n+1)π - 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 = 2nπ + = 2nπ - -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:
We have tan2x = Illustrative problem Solve tan2x = Solution:
Illustrative problem Solve sin2x + sin4x + sin6x = 0 Solution:
Illustrative Problem Solve 2cos2x + 3sinx = 0 Solution:
General solution of sin2x = sin2cos2x = 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:
Solve : Class Exercise Q4. Solution:
Class Exercise Q5. Show that 2cos2(x/2)sin2x = x2+x-2 for 0<x</2 has no real solution. Solution:
Find the value(s) of x in (- , ) which satisfy the following equation Class Exercise Q6. Solution:
Find the value(s) of x in (- , ) which satisfy the following equation Class Exercise Q6. Solution:
Class Exercise Q7. Solve the equation sinx + cosx = 1+sinxcosx Solution:
Class Exercise Q7. Solve the equation sinx + cosx = 1+sinxcosx Solution:
If rsinx=3,r=4(1+sinx), then x is Class Exercise Q8. Solution:
Class Exercise Q8. If rsinx=3,r=4(1+sinx), then x is Solution:
In a ABC , A > B and if A and B satisfy 3 sin x –4 sin3x – k = 0 ( 0< |k| < 1 ) , C is Class Exercise Q9. Solution:
Class Exercise Q10. Solve the equation (1-tan)(1+sin2) = 1+tan Solution:
Class Exercise Q10. Solve the equation (1-tan)(1+sin2) = 1+tan Solution:
Class Exercise Q10. Solve the equation (1-tan)(1+sin2) = 1+tan Solution: