Solutions of Trigonometric Equations

For a Given Interval, Solve the Trigonometric Equations

Solutions of Trigonometric Equations for a Given Interval

Edexcel Syllabus stipulates:

“Students should be able to solve equations such as:

1)    sin (x – π/2) = ¾,  for 0 < x < 2π

2)    cos (x + 30^o) = ½, for -180^o < x < 180^o

3)    tan 2x = 1,  for 90^o < x < 270^o

4)    6 cos² x^o  + sin x^o – 5 = 0, for 0  ≤ x < 360

5)    sin² (x + π/6) = ½,  for –π ≤ x < π.

6)   Past Year Question: Edexcel Pure Maths 7362/02, Q8 (15 May 2008)

Solve, to 3 significant figures, for 0 ≤ Ө ≤ π,

(a)  (4 sin Ө – 1)(2 sin Ө + 5) = 0                                                       (3)

(b)  tan(2Ө – π/3) = 2.4                                                                       (4)

(c)  9 sin² Ө – 9 cos Ө = 11                                                                (5)

7) Solve,   4 sin Ө cos Ө – 10 sin Ө - 2 cos Ө + 5 = 0, for - 90 ≤ Ө ≤ 90 (in degrees)

     

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1)      Solve sin (x – π/2) = ¾,  for 0 < x < 2π

To solve the trigo. equation is to find the value(s) of x for the given interval or domain: 0 < x < 2π

           

            To find x, you have to use the given relationship: sin (x – π/2) = ¾

Based on this relationship:

·           The Reference Angle, R = sin^-1 (3/4)

·           The Principle Angle P (x – π/2) is in

o    The Principle Domain: –π/2  < P < 2π – π/2

o    1^st quadrant (where, P = R): (x – π/2) = sin^-1 (3/4),

o    2^nd quadrant (where P = π – R): (x – π/2) = π - sin^-1 (3/4),

Hence,

1^st quadrant answer:

(x – π/2) = sin^-1 (3/4)

            x = π/2 + sin^-1 (3/4)

            x = 2.418 858 406

            x = 2.42 (to 3 s.f.)

                       

                        2^nd  quadrant answer:

                        (x – π/2) = π - sin^-1 (3/4)

                                    x = π + π/2 - sin^-1 (3/4)

                                    x = 3.864 326 901

                                    x = 3.86 (to 3 s.f.)

2)      Solve cos (x + 30^o) = ½, for -180^o < x < 180^o

Let:      R be the Reference Angle: R = cos^-1 (1/2)

P be (x + 30^o).

Therefore:

Domain for P: -150^o < P < 210^o

(since, -180^o + 30^o < (x + 30^o) or P < 180^o+ 30^o)

P (or, x + 30^o) must be in 1^st Qd or 4^th Qd

1^st Qd: P = R = cos^-1 (1/2) = 60^o

            x + 30^o = 60^o

                        x = 30^o

4^th Qd:             P = -60^o

              x + 30^o = -60^o

                        x = -90^o

6)    Past Year Question: Edexcel Pure Maths 7362/02, Q8 (15 May 2008)

Solve, to 3 significant figures, for 0 ≤ Ө ≤ π,

(a)  (4 sin Ө – 1)(2 sin Ө + 5) = 0                                                       (3)

(b)  tan(2Ө – π/3) = 2.4                                                                       (4)

(c)  9 sin² Ө – 9 cos Ө = 11                                                                (5)

Solutions:

(a)    (4 sin Ө – 1)(2 sin Ө + 5) = 0

(4 sin Ө – 1) = 0;                          or, (2 sin Ө + 5) = 0

sin Ө = ¼                                     or, sin Ө = -5/2

      Ө = sin^-1 (1/4)                        but, sin must be ≤ 1

      Ө = 0.252 680 255                 so, (2 sin Ө + 5) ≠ 0

     

      Ө = 0.253 (to 3 s.f.)

(b)    tan(2Ө – π/3) = 2.4

Reference Angle R = tan^-1 (2.4) 
                              ( = 1.176005207 rad. - but NO need to get. this value early)

Let P be (2Ө – π/3). Therefore,

Domain for P is: – π/3 ≤ P ≤ 2π – π/3

                      or, – π/3 ≤ P ≤ 5π/3

                    

For P’s domain, since tan = +ve value,

P must be in 1^st Quadrant;           or, 3^rd Quadrant:

P = R = tan^-1 (2.4);                       or, P = π + R = π + tan^-1 (2.4)

(2Ө – π/3) = tan^-1 (2.4)                or, (2Ө – π/3) = π + tan^-1 (2.4)

 2Ө = π/3 + tan^-1 (2.4)                 or, 2Ө = 4π/3 + tan^-1 (2.4)

   Ө = [π/3 + tan^-1 (2.4)] ÷ 2        or,  Ө = [4π/3 + tan^-1 (2.4)] ÷ 2                             

   Ө = 1.111 601 379                    or 2.682 397 706

 

  Ө = 1.11 or 2.68 (to 3 s.f.)

(c)       9 sin² Ө – 9 cos Ө = 11

9 (1 – cos² Ө) – 9 cos Ө = 11 ….(using: cos² Ө + sin² Ө = 1)

9 –9 cos² Ө – 9 cos Ө = 11

9 - 11 – 9 cos² Ө – 9 cos Ө = 0

X (-1): 9 cos² Ө + 9 cos Ө + 2 = 0

(3 cos Ө + 1)( 3 cos Ө +2) = 0

(3 cos Ө + 1) = 0                                   or, (3 cos Ө + 2) = 0

cos Ө = - (1/3)                                       or, cos Ө = - (2/3)

Given the domain for Ө:  0 ≤ Ө ≤ π, and cos Ө = -ve values

Ө values must be in the 2^nd Qd.

Hence, Ө = π – cos^-1 (1/3)        or, = π – cos^-1 (2/3)

          Ө = 1.910 633 236           or, = 2.300 523 983

Ө = 1.91            or, 2.30 (to 3 s.f.)

7) Solve,  4 sin Ө cos Ө – 10  sinӨ - 2 cos Ө + 5 = 0, for - 90 ≤ Ө ≤ 90 (in degrees)

            Factorise:         2 cos Ө (2 sin Ө – 1) – 5 (2 sinӨ – 1) = 0

                                    (2 sinӨ – 1)( 2 cosӨ – 5) = 0

            Imply:               (2 sinӨ – 1) = 0            or         (2 cosӨ – 5) = 0

                                    sin Ө = ½                      or         cos Ө = 5/2

                                    Ө = sin^-1(1/2)               but       cos Ө ≤ 1

                                    Ө = 30^o                        so,        (2 cos Ө – 5) ≠ 0

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Your Homework:

3)    Solve, tan 2x = 1,  for 90^o < x < 270^o

4)    Solve, 6 cos² x^o  + sin x^o – 5 = 0, for 0  ≤ x < 360

5)    Solve, sin² (x + π/6) = ½,  for –π ≤ x < π.   

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SPM 2013 Paper 2 Q 4(b) on Solutions of Trigonometric Equation:

4.         (a) Prove that tan x sin 2x = 1 – cos 2x             (2)

            (b) Hence, solve the equation

tan x sin 2x = ¼, for 0^o ≤ x ≤ 360^o                    (4)

Proposed Solutions

4          (a) To prove: tan x sin 2x = 1 – cos 2x

                        LHS = tan x sin 2x

                        Use identities:

                        tan x = sin x / cos x; and

                        sin 2x = 2 sin x cos x

                        LHS = (sin x / cos x)( 2 sin x cos x) = 2 sin² x

                        Use identities:

cos 2x = 1 - 2 sin² x; or,

2 sin²x = 1 – cos 2x

Therefore,

LHS = 2 sin²x = 1 – cos 2x = RHS (Proven)

            (b) Hence, tan x sin 2x = ¼, for 0^o ≤ x ≤ 360^o

                        Imply: 1 – cos 2x = ¼, for 0^o ≤ 2x ≤ 720^o

                                    cos 2x = 1 – ¼ = ¾ (+ve)

                       

Thus, for domain of 2x: 0^o ≤ 2x ≤ 720^o, 2x^o must be in

·        1^st Quadrant:

P₁ = 2x^o = Reference Angle (R) = cos^-1 (3/4)

x^o = (1/2)[cos^-1 (3/4)] = 20.70481106…^o

x = 20.7 (to 1 dp); or

(1^st Qd 1st co-terminal angle)

2x = 360(1) + P₁ = 360 + cos^-1 (3/4)

x = (1/2)[360 + cos^-1 (3/4)] =  200.7048111…

   = 200.7 (to 1 dp)

·        4^th Quadrant:

P₂ = 2x^o = 360^o – R = 360^o – cos^-1 (3/4)

x^o = (1/2)[ 360^o  - cos^-1 (3/4)] = 159.2951889…^o

x = 159.3 (to 1 dp); or

(4^th Qd 1st co-terminal angle)

2x^o = 360^o + P₂ = 360^o + [360^o – cos^-1 (3/4)]

x^o = (1/2)[720^o – cos^-1 (3/4)]

x = 339.2951889

x = 339.3 (to 1 dp)

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