Mathematics

Mathematics

Trigonometric Equations – Session 2

Session Objectives

Session Objectives

Removing Extraneous Roots

Avoiding Root Loss

Equation of the form of a cosx +b sinx = c

Simultaneous equation

Extraneous solutions

The solutions, which do not satisfy the trigonometric equation.

Trigonometric Equation – Removing Extraneous Solutions

Origin of Extraneous solutions ?

Squaring during solving the trigo. equations. ( ±)2= +

Solutions, which makes the equation undefined. ( denominator being zero for a set of solutions) Esp . equations containing tan or sec, cot or cosec

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Trigonometric Equation – Extra root because of squaring

Illustrative problem

Solve sec - 1 = ( √ 2 – 1) tan

Equation can be rewritten as sec = ( √ 2 – 1) tan + 1

On squaring , we get

sec2 = (2+1–2√2)tan2 +1+2(√2–1)tan

sec2 - tan2 = (2–2√2)tan2 +1 + 2(√2–1)tan

(2 – 2√2 ) tan2 + 2(√2 – 1) tan = 0

tan = 0 and tan = +1

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Trigonometric Equation – Extra root because of squaring

Solve sec - 1 = ( √ 2 – 1) tan

tan = 0 or tan = +1

= n or = n + /4

Putting = in given equation , we get

L.H.S. = ( -1 –1) = -2

R.H.S. = ( √ 2 – 1).0 = 0

Extraneous solutions : = odd integer multiple of π

WHY ??Answer : = 2n or = 2n + /4

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Trigonometric Equation – Solutions, which makes the equation undefined

Solve tan5 = tan3

Solution will be 5 = nπ + 3

= nπ/2 , where n Z

solutions = π/2, 3π/2…..etc. will make tan3 and tan5 undefined

Answer : = mπ, where m Z

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Trigonometric Equation – Avoiding root loss

Reason for root loss ?

Canceling the terms from both the sides of the equation

Use of trig. Relationship , which restricts the acceptable values of ,i.e., the domain of changes.

2tan1

2tan1

cos2

2

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Trigonometric Equation – Cancelling of terms from both sides

Illustrative problem

Solve sin .cos = sin

If we cancel sin from both the sides

cos = 1 = 2n , where n Z

However , missed the solution provided by sin = 0

= n ,where n Z

Answer : = n ,where n Z

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Trigonometric Equation – Changes in domain of

Illustrative Problem

Solve : sin - 2.cos = 2 If we use

2tan1

2tan1

cos2

2

2tan12

tan2sin

2

2

2tan1

2tan1

.2

2tan12

tan2

2

2

2

Equation can be written as :

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Trigonometric Equation – Changes in domain of

Solve : sin - 2.cos = 2

2

21

21

2

21

22

2

2

2

tan

tan.

tan

tan

2tan22

2tan22

2tan2 22

22

tan = 2nπ + 2 , where n Z and = tan-1 2

Root loss : = (2n+ 1) π where , n Z

As = π , 3 π , 5 π …. satisfy the given trig. equation

Answer : = (2nπ+2) U (2n+1)π, where n Z and = tan-1 2

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Trigonometric Equation – a sin + b cos = c

Reformat the equation in the form of

cos( - ) = k.

1. Divide both sides of the equation by

2b2a

2. The equation now will be as

2b2a

ccos2b2a

bsin2b2a

a

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Trigonometric Equation – a sin + b cos = c

2b2a

ccos2b2a

bsin2b2a

a

Compare with sin.sin + cos.cos = k

2b2a

asin

2b2a

bcos

2b2a a

b

Hence , the given equation can be written as

ba1tanwhere,

2b2a

c)(cos

For real values of , 12b2a

c1

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Trigonometric Equation – a sin + b cos = c - Algorithm

Step 2:

Check whether real solution exists

12b2a

c1

Step 1:

Reformat the equation into cos(-) = k

Step 3:

Solve the equation cos(-) = k

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Solve sin + cos = 1

Trigonometric Equation – a sin + b cos = c - Problem

Step 1:

Reformat the equation into cos(-) = k

2121

1cos2121

1sin2121

1

21cos

21sin

21

21

4cos

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Solve sin + cos = 1


1,1inis21As

Step 2:

Check whether real solution exists

12b2a

c1

Real solution exists

21

4cos

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Solve sin + cos = 1


Step 3:

Solve the equation cos(-) = k

21

4cos

4cos

4cos

4n2

4

2n2orn2

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Simultaneous Trigonometric Equations

Case I : Two equations and one variable angle

Step 1 – Solve both the equation between 0 and 2π .

Step 2 – Find common solutions.

Step 3– Generalise the solution by adding 2nπ to common solution as per step 2.

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Find the general solution of tan = -1, cos = 1/2

47,

43,1tanFor

Step 1 – Solve both the equation between 0 and 2π

47,

4,2

1cosFor

Simultaneous Trigonometric Equations - Problem _J33


Find the general solution of tan = -1, cos = 1/2

Simultaneous Trigonometric Equations - Problem

4

7

Step 2 – Find common solutions

Step 3– Generalise the solution 47n2

Answer

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Case II :

Two equations and two variable angles(θ,φ) and smallest positive values of the angles satisfying the equations needs to be found out

Find the smallest positive values of θ and φ , if tan(θ – φ) = 1 and sec ( θ+φ) = 2/ √3

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Step 1 – Solve both the equations between 0 and 2π .

Step 2 –

Find two equations with the conditions such as ( θ+φ) > (θ-φ) for positive θ and φ

Step 3 - Solve the two equations to determine θ and φ

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Step 1 –

Solve both the equations between 0 and 2π .

45,

4

611,

6

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45,

4 6

11,6

Step 2 – Find two equations with the conditions such as ( θ+φ) > (θ-φ) for positive θ and φ

4611

Step 3 - Solve the two equations to determine θ and φ

2419;

2425

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Trigonometric Equation – Misc. Tips :

1. cosθ = k is simpler to solve compared to sinθ = k

2. Check for extraneous roots and root loss

3. In case of ‘algebraic function of angle’ is a part of the equation use of the following properties:

• Range of values of sin and cos functions

• x2 ≥ 0

• A.M. ≥ G.M.

xx2

2 226xx

cos2


4. Equation with multiple terms of the form (sinθ ± cosθ) and sinθ.cosθ :

• Put sinθ+cosθ = t.

• Equation gets converted in to a quadratic equation in t

sin x + cos x = 1+ sin x. cos x


5. Equation with terms of sin2θ , cos2θ and sinθ.cosθ

• Try dividing by cos2θ to get a quadratic equation in tan θ.

6. Solution to sin2θ = sin2 , cos2θ = cos2 and tan2θ = tan2 is n

Solve the equation 2 sin2 x – 5 sinx.cos x – 8 cos2 x = -2

Class exercise Q1

Number of solution of the equation tanx+secx = 2cosx lying in the interval [0,2]

(a) 2 (b) 3 (c ) 0 (d) 1

Solution:

tanx secx 2cosx wherex [0,2 ]

2sinx 1 2cos x

22 2sin x sinx 1

cos x 0

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Class exercise Q1


22sin x sinx 1 0 22sin x 2sinx sinx 1 0

2sinx(sinx 1) 1(sinx 1) 0 1sinx or sinx 1

2

n nx n ( 1) or x n ( 1) ( )6 2

for x [0,2 ] n 0 x or x not valid6 2

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Class exercise Q1


5 3n 1 x or x

6 2

13 3

n 2 x ( 2 ) or x6 2

5 3x , ,

6 6 2

3however for x , tanx and secx is undefined.

2

5hence solution in [0,2 ] is x ,

6 6

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Class exercise Q2

I f 2 secx tanx 1, x is

(a)n (b)n (c)2n c none4 4 4

2 22sec x 1 tan x 2tanx 2 22 2tan x 1 tan x 2tanx

x n4

2 secx (1 tanx)

2tan x 2tanx 1 0 2(tanx 1) 0 tanx 1

Solution:

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Class exercise Q2

I f 2 secx tanx 1, x is

As we have squared both the sides, we should check for extraneous roots

n 0, x4

2

L.H.S 2. 1 2 1 1 R.H.S1

3n 1, x

4

L.H.S 2. 2 1 2 1 3 R.H.S

Similarly, for n=1, 3, 5 …the values of x do not satisfy the question.

Hence, the solution is x 2n4

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Class exercise Q3

For nz, the general solution of the equation

3 1 sinx 3 1 cosx 2 is

n

n

(a) x n 14 12

(c) x 2n 14 12

(b) x 2n4 12

(d) x 2n4 12

Solution:

3 1 sinx 3 1 cosx 2

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Class exercise Q3

For nz, the general solution of the equation

3 1 sinx 3 1 cosx 2 is

3 1 sinx 3 1 2cosx

2 2 2 2 2 2

cos x cos12 4

x 2n4 12

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Class exercise Q4

1sin(x y) cos(x y) ,

2

the values of x and y lying between 0o and 90o are given by

(a) x=15o , y=25o (b) x=65o , y=15o

(c) x=45o , y=45o (d) x=45o , y=15o

Solution:1

if sin(x y)2

5x y ,

6 6

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Class exercise Q4

1cos (x y)

2

5x y '

3 3

for x and y 0,2

x y6

2x x 452 4

x y y 306

(x y)3

1sin(x y) cos(x y) ,

2

the values of x and y lying between 0o and 90o are given by

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Class exercise Q5

2 2

2

Solve (sinx cosx) (sinx 1)

(cosx 1) 0

Solution:

L.H.S =0 if sinx-cosx=0, sinx-1=0 cosx-1=0

sinx=cosx, sinx=1, cosx=1

No Solution

Which is not possible for any value of x.

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Class exercise Q6

I f cot x cosecx 3, then x is

(a) 2n (b) n3 3

(c) n (d) Noneof these3

Solution:

given cot x cosecx 3

cosecx 3 cot x 2 2cosec x 3 cot x 2 3 cot x (squaring)

2 2 3 cot x 0

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Class exercise Q6

1cot x

3

x n3

for x , .....LHS RHS3 3

Hence acceptable solution is x 2n3

I f cot x cosecx 3, then x is _J30

Class exercise Q7

Solve sinx 3 cosx 2

Solution:the given equation of the form a cos + b sin = c

Here a 3,b 1, c 2

for real solution2 2| c | a b

i.e| 2 | 3 1

i.e| 2 | 2 whish is true

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Class exercise Q7

Solve sinx 3 cosx 2

2 2

Hence dividing both side by

a b i.e. 2, we get

3 1 1cos x sinx

2 2 2

cosx.cos sin x.sin cos6 6 4

cos x cos6 4

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Class exercise Q7

Solve sinx 3 cosx 2

x 2n6 4

5x 2n 2n

4 6 12

x 2n6 4

Taking positive sign

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Class exercise Q7

Solve sinx 3 cosx 2

x 2n6 4

x 2n 2n4 6 12

5x 2n , 2n where n I

12 12

Taking Negative sign

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Class exercise Q8

2 2

Solve the equation

2sin x 5sinx cosx 8cos x 2

Solution:

In such types of problems we divide both sides by cos2x which yield a quadratic equation in tanx.

povided cosx 0 i.e. x2

In this equation if cosx = 0, the equation becomes 2sin2x=-2 or sin2x=-1 which is not possible

hence on dividing the equation by cos2x we get

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Class exercise Q8

2 2

Solve the equation


2tan2x-5tanx-8 = -2sec2x

2tan2x+2(1+tan2x)-5tanx-8 = 0

or 4tan2x-5tanx-6 = 0

or 4z2-5z-6 = 0 where z = tanx

or 4z2-8z+3z-6 = 0

4z(z-2)+3(z-2)=0 z=2,-3/4

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Class exercise Q8

2 2

Solve the equation


3i.e tanx 2; tanx

4

1 1 3x tan 2; x tan

4

1 1 3x n tan 2; x n tan ,n I

4

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Class exercise Q9

Determine for which value of ‘a’ the equation a2– 2a+sec2(a+x)=0 has solution and find the solution

The equation involves two unknown a and x so we must get two condition for determining unknowns since R.H.S is zero. So break the L.H.S of the equation as sum of two square.

2 2a 2a sec (a x) 0

2 2or a 2a 1 tan (a x) 0

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Class exercise Q9

Determine for which value of ‘a’ the equation a2– 2a+sec2(a+x)=0 has solution and find the solution

2 2(a 1) tan (a x) 0

2 2(a 1) 0, tan (a x) 0

a=1 and tan(a+x)=0 but a=1

(1 x) n x n (n 1)

x n 1 where n I x m where m I

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Class exercise Q10

Solve cot – tan = sec

cos sin 1 cos2 1sin cos cos sin cos cos

21 2sin sin (cos 0)

22sin sin 1 0

(2sin 1)(sin 1) 0

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Class exercise Q10

Solve cot – tan =sec

n1sin n ( 1)

2 6

But sin 1 sec is undefined

sin 1

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