Solve the equation cos 2 θ = 5 + 1 4 , θ ∈ [ 0 , 2 π ] .

Given cos 2 θ = 5 + 1 4 , θ ∈ [ 0 , 2 π ] We know that cos 36 ° = 5 + 1 4 Therefore, cos 2 θ = cos 36 ° = cos π 5 For any real numbers x and y , cos x = cos y , implies x = 2 n π ± y , where n ∈ Z Therefore 2 θ = 2 n π ± α ⇒ 2 θ = 2 n π ± π 5 , ⇒ θ = n π ± π 10 , θ ∈ [ 0 , 2 π ] Hence, θ = π 10 , 9 π 10 , 11 π 10 , 19 π 10

Solve the equation tan 2 θ = 1 , θ ∈ [ - π , π ] .

Given tan 2 θ = 1 , θ ∈ [ - π , π ] ⇒ tan θ = ± 1 We know that tan π 4 = 1 and tan - π 4 = 1 θ = n π + α θ = n π + π 4 or θ = n π - π 4 where n ∈ Z and θ ∈ [ - π , π ] Therefore θ = ± π 4 , ± 3 π 4

Solve the equation sin 3 θ = 3 2 , θ ∈ [ - π , π ] .

Given sin 3 θ = 3 2 We know that sin π 3 = 3 2 Therefore, sin 3 θ = sin π 3 For any real numbers x and y , sin x = sin y , implies x = n π + - 1 n y , where n ∈ Z Therefore, 3 θ = n π + - 1 n π 3 θ = n π 3 + - 1 n π 6 and θ ∈ [ - π , π ] n = 1 ⇒ θ = π 3 - π 9 = 2 π 9 h = 2 ⇒ θ = 2 π 3 + π 9 = 7 π 9 n = 3 ⇒ θ = π - π 9 = 8 π 9 n = - 1 ⇒ θ = - π 3 - π 9 = - 4 π 9 n = - 2 ⇒ θ = - 2 π 3 + π 9 = - 5 π 9 n = 0 ⇒ θ = π 9 Hence, θ = - 5 π 9 , - 4 π 9 , π 9 , 2 π 9 , 7 π 9 , 8 π 9

Solve the equation cos 2 θ = 3 4 · θ ∈ [ 0 , π ] .

Given cos 2 θ = 3 4 , θ ∈ [ 0 , π ] ⇒ cos 2 θ = 3 4 = 3 2 2 ⇒ cos 2 θ = cos 2 π 6 If cos 2 θ = cos 2 α then θ = n π ± α , n ∈ Z θ = n π ± π 6 For n = 0 ⇒ θ = π / 6 n = 1 ⇒ θ = π - π / 6 , π + π / 6 Hence, θ = π 6 , 5 π 6

Solve the equation 2 sin 2 θ = sin θ , θ ∈ ( 0 , π ) .

Given 2 sin 2 θ = sin θ , θ ∈ ( 0 , π ) ⇒ 2 sin 2 θ - sin θ = 0 ⇒ sin θ 2 sin θ - 1 = 0 ⇒ sin θ = 0 or 2 sin θ - 1 = 0 Therefore, sin θ = sin n π , Hence , θ = n π Now, sin θ = 1 2 = sin π 6 , ⇒ θ = n π + - 1 n π 6 and θ ∈ ( 0 , π ) For n = 1 ⇒ θ = π - π 6 = 5 π 6 Hence, θ = π 6 , 5 π 6

E M B I B E

Intermediate First Year Mathematics Paper 1A>Algebra, Vector Algebra, Trigonometry>Chapter 7 - Trigonometric Equations>Exercise 7 (a)

Andhra Pradesh Board Solutions for Exercise 1: Exercise 7 (a)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 1: Exercise 7 (a)

Attempt the free practice questions from Exercise 1: Exercise 7 (a) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1A solutions are prepared by Experienced Embibe Experts.