Solve the given equation for - 180 ° ≤ x ≤ 180 ° . cos 2 x = 3 2

Given equation is cos 2 x = 3 2 ⇒ cos 2 x = cos 30 ° . . . i We know that the general solution of the equation cos θ = cos α is given by θ = 360 ° n ± α , n ∈ Z Thus, the general solution of equation i is given by 2 x = 360 ° n ± 30 ° ⇒ x = 180 ° n ± 15 ° . . . i i But we need to find the values of x for - 180 ° ≤ x ≤ 180 ° . Thus, we substitute n = 0 in equation i i ⇒ x = ± 15 ° . . . . i i i For n = 1 , from equation i i , we have x = 180 ° - 15 ° = 165 ° . . . . i v [we will not consider x = 180 ° + 15 ° , as - 180 ° ≤ x ≤ 180 ° ] And, for n = - 1 from equation i i , we have x = - 180 ° + 15 ° = - 165 ° . . . . v [again, we will not consider x = - 180 ° - 15 ° as - 180 ° ≤ x ≤ 180 ° ] Hence, from i i i , i v , and v The required values of x are - 15 ° , 15 ° , - 165 ° , and 165 ° .

Solve the given equation for - 180 ° ≤ x ≤ 180 ° . cos 3 x = 3 2

Given equation is cos 3 x = 3 2 ⇒ cos 3 x = cos 30 ° . . . i We know that the general solution of the equation cos θ = cos α is given by θ = 360 ° n ± α , n ∈ Z Thus, the general solution of equation i is given by 3 x = 360 ° n ± 30 ° ⇒ x = 120 ° n ± 10 ° . . . i i But we need to find the values of x for - 180 ° ≤ x ≤ 180 ° . Thus, we substitute n = 0 in equation i i ⇒ x = ± 10 ° . . . . i i i For n = 1 , from equation i i , we have x = 120 ° ± 10 ° ⇒ x = 120 ° + 10 ° = 130 ° and x = 120 ° - 10 ° = 110 ° . . . . i v For n = - 1 from equation i i , we have x = 120 ° - 1 ± 10 ° ⇒ x = - 120 ° + 10 ° = - 110 ° , and x = - 120 ° - 10 ° = - 130 ° . . . . v Hence, the solution of the given equation for - 180 ° ≤ x ≤ 180 ° are ± 10 ° , 110 ° , 130 ° , - 110 ° , and - 130 °

Solve the given equation for - 180 ° ≤ x ≤ 180 ° . 2 cos 3 x - 1 = 0

Given equation is 2 cos 3 x - 1 = 0 ⇒ cos 3 x = 1 2 ⇒ cos 3 x = cos 60 ° . . . . i We know that the general solution of the equation cos θ = cos α is given by θ = 360 ° n ± α , n ∈ Z Thus, the general solution of equation i is given by 3 x = 360 ° n ± 60 ° ⇒ x = 120 ° n ± 20 ° . . . i i But we need to find the values of x for - 180 ° ≤ x ≤ 180 ° . Thus, we substitute n = 0 in equation i i ⇒ x = ± 20 ° . . . . i i i For n = 1 , from equation i i , we have x = 120 ° ± 20 ° ⇒ x = 120 ° + 20 ° = 140 ° and x = 120 ° - 20 ° = 100 ° . . . . i v For n = - 1 from equation i i , we have x = 120 ° - 1 ± 20 ° ⇒ x = - 120 ° + 20 ° = - 100 ° , and x = - 120 ° - 20 ° = - 140 ° . . . . v Hence, the solution of the given equation for - 180 ° ≤ x ≤ 180 ° are ± 20 ° , 140 ° , 100 ° , - 100 ° , - 140 ° .

Solve the given equation for - 180 ° ≤ x ≤ 180 ° . 3 tan x 2 + 3 = 0

Given equation is 3 tan x 2 + 3 = 0 ⇒ tan x 2 = - 1 ⇒ tan x 2 = tan 180 ° - 45 ° = tan 135 ° . . . . i We know that the general solution of the equation tan θ = tan α is θ = 180 ° n + α , n ∈ Z . Thus, the general solution of equation i is given by x 2 = n 180 ° + 135 ° , n ∈ Z . . . . ii For n = - 1 , from equation i i , we have x 2 = - 180 ° + 135 ° = - 45 ° ⇒ x = - 90 ° For all other values of n , from equation i i , we get those values of x for which x ∉ - 180 ° , 180 ° Hence, the required solution is x = - 90 °

Solve the given equation for 0 ≤ θ ≤ 2 π . sin 3 θ = 3 2

Given equation is sin 3 θ = 3 2 ⇒ sin 3 θ = sin π 3 . . . . i We know that the general solution of the equation sin θ = sin α is θ = n π + ( - 1 ) n α , n ∈ Z . Thus, the general solution of equation i is given by 3 θ = n π + - 1 n π 3 , n ∈ Z ⇒ θ = n π 3 + - 1 n π 9 , n ∈ Z . . . . i i Substituting n = 0 in equation i i , we have θ = 0 + - 1 0 π 9 = π 9 Substituting n = 1 in equation i i , we have θ = π 3 - π 9 = 3 π - π 9 = 2 π 9 Substituting n = 2 in equation i i , we have θ = 2 π 3 + - 1 2 π 9 = 2 π 3 + π 9 = 7 π 9 For n = 3 , from equation i i , we have θ = π - π 9 = 8 π 9 For n = 4 from equation i i , we have θ = 4 π 3 + - 1 4 π 9 = 4 π 3 + π 9 = 13 π 9 For n = 5 from equation i i , we have θ = 5 π 3 + - 1 5 π 9 = 5 π 3 - π 9 = 14 π 9 For n = 6 from equation i i , we have θ = 6 π 3 + - 1 6 π 9 = 2 π + π 9 = 19 π 9 > 2 π , and does not lie in the given range. i.e., 0 ≤ θ ≤ 2 π . Hence, the required solutions of the given equation for 0 ≤ θ ≤ 2 π are π 9 , 2 π 9 , 7 π 9 , 8 π 9 , 13 π 9 , 14 π 9 .

Solve the given equation for 0 ≤ θ ≤ 2 π . cos 3 θ - 1 = 0

Given equation is cos 3 θ - 1 = 0 ⇒ cos 3 θ = 1 ⇒ cos 3 θ = cos 0 ° . . . . i We know that the general solution of the equation cos θ = cos α is θ = 2 n π ± α , n ∈ Z . Thus, the general solution of equation i is given by 3 θ = 2 n π ± 0 , n ∈ Z ⇒ θ = 2 nπ 3 , n ∈ Z . . . . . i i Now, we need to find those solutions for which 0 ≤ θ ≤ 2 π . We substitute n = 0 in equation i i ⇒ θ = 0 For n = 1 , we have θ = 2 π 3 For n = 2 , we have θ = 4 π 3 For n = 3 , we have θ = 6 π 3 = 2 π Hence, the solutions of the given equation for 0 ≤ θ ≤ 2 π are 0 , 2 π 3 , 4 π 3 , and 2 π

Solve the given equation for 0 ≤ θ ≤ 2 π . sin θ 2 = 2 2

Given equation is sin θ 2 = 2 2 ⇒ sin θ 2 = sin 1 2 ⇒ sin θ 2 = sin π 4 . . . . i We know that the general solution of the equation sin x = sin y is x = n π + - 1 n y , n ∈ Z . Thus, the general solution of equation i is given by θ 2 = n π + - 1 n π 4 , n ∈ Z . . . . . i i Now, we need to find those solutions of the given equation for which 0 ≤ θ ≤ 2 π . Thus, using equation i i For n = 0 , we have ⇒ θ 2 = 0 + - 1 0 π 4 = π 4 ⇒ θ = π 2 For n = 1 , we have θ 2 = π - π 4 = 3 π 4 ⇒ θ = 3 π 2 For any other value of n , we will observe that θ ∉ 0 , 2 π . Hence, the required solutions of the given trigonometric equations are π 2 , and 3 π 2

Solve the given equation for 0 ≤ θ ≤ 2 π . sin 2 2 θ 3 - 1 = 0

Given equation is sin 2 2 θ 3 - 1 = 0 ⇒ sin 2 2 θ 3 = 1 ⇒ sin 2 θ 3 = ± 1 ⇒ sin 2 θ 3 = 1 and sin 2 θ 3 = - 1 ⇒ sin 2 θ 3 = sin π 2 . . . . i , and sin 2 θ 3 = sin π + π 2 = sin 3 π 2 . . . . i i We know that the general solution of the equation sin x = sin y is x = n π + - 1 n y , n ∈ Z . Thus, the general solution of equation i is given by 2 θ 3 = n π + - 1 n π 2 . . . . i i i And, the general solution of equation i i is given by 2 θ 3 = n π + - 1 n 3 π 2 . . . . i v For n = 0 from equation i i i , we have 2 θ 3 = π 2 ⇒ θ = 3 π 4 . . . v And, for n = 0 from equation i v , we have 2 θ 3 = 3 π 2 ⇒ θ = 9 π 4 , which we don't have to consider as 0 ≤ θ ≤ 2 π . For n = 1 , from equation i i i , we have 2 θ 3 = π - π 2 = π 2 ⇒ θ = 3 π 4 . . . . . v i Thus, the required solution of the given equation for which 0 ≤ θ ≤ 2 π is θ = 3 π 4 .

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12E

Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 16: Exercise 12E

Author:Natasha Awada, Paul La Rondie, Laurie Buchanan & Jill Stevens

Natasha Awada Mathematics Solutions for Exercise - Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 16: Exercise 12E

Attempt the practice questions on Chapter 12: Periodic Relationships: Trigonometric Functions, Exercise 16: Exercise 12E with hints and solutions to strengthen your understanding. Mathematics : Analysis and Approaches Standard Level Course Companion solutions are prepared by Experienced Embibe Experts.