Solve the equation 2 sin 2 x - cos x - 1 = 0 for 0 ∘ ⩽ x ⩽ 360 ∘ .

Given that, 2 sin 2 x - cos x - 1 = 0 , 0 ⩽ x ≤ 360 ° ∴ 2 sin 2 x - cos x - 1 = 0 As sin 2 x + cos 2 x = 1 ∴ 2 1 - cos 2 x - cos x - 1 = 0 ∴ 2 - 2 cos 2 x - cos x - 1 = 0 ∴ 2 - 2 cos 2 x - cos x - 1 = 0 ∴ - 2 cos 2 x - cos x + 1 = 0 ∴ 2 cos 2 x + cos x - 1 = 0 ∴ 2 cos 2 x + 2 cos x - cos x - 1 = 0 ∴ 2 cos x cos x + 1 - 1 cos x + 1 = 0 ∴ 2 cos x - 1 cos x + 1 = 0 Now, ∴ cos x = - 1 ∴ cos x = cos 180 ° ∴ x = 180 ° Also, cos x = 1 2 ∴ cos x = cos π 3 ------- cos π 3 = 1 2 ∴ x = π 3 ∴ x = 60 Again we know that, cos x = 1 2 cos x = cos 5 π 3 ------------ cos 5 π 3 = 1 2 ∴ x = 5 π 3 ∴ x = 300

Solve the equation 3 cos 2 x - 3 = sin x for 0 ∘ ⩽ x ⩽ 360 ∘ .

Given that, 3 cos 2 x - 3 = sin x , 0 ⩽ x ⩽ 360 ° ∴ 3 1 - sin 2 x - 3 - sin x = 0 ∴ 3 - 3 sin 2 x - 3 - sin x = 0 ∴ 3 sin 2 x + sin x = 0 ∴ sin x ( 3 sin x + 1 ) = 0 ∴ sin x = 0 or 3 sin x + 1 = 0 ∴ sin x = 0 or sin x = - 1 3 Let, ∴ sin x = 0 ∴ x = π , 2 π ∴ x = 180 ° , 360 ° Also, sin x = - 1 3 ∴ x = sin - 1 - 1 3 ∴ x = 199 . 268 °

Solve the equation cos x + 5 = 6 sin 2 x for 0 ∘ ⩽ x ⩽ 360 ∘ .

cos x + 5 = 6 sin 2 x for 0 ° ⩽ x ⩽ 360 ° 6 sin 2 x - cos x - 5 = 0 Using the identity: sin 2 x + cos 2 x ≡ 1 and rearranging: sin 2 x ≡ 1 - cos 2 x multiplying by 6 gives: 6 sin 2 x ≡ 6 - 6 cos 2 x then substituting gives: 6 - 6 cos 2 x - cos x - 5 = 0 6 cos 2 x + cos x - 1 = 0 Factorising: ( 3 cos x - 1 ) ( 2 cos x + 1 ) = 0 * Either: 3 cos x - 1 = 0 or 2 cos x + 1 = 0 cos x = 1 3 or cos x = - 1 2 If cos x = 1 3 x = 70 . 52 … ° or 360 ° - 70 . 52 … ° = 289 . 47 ° Or: 2 cos x + 1 = 0 cos x = - 1 2 x = 120 ° , 240 ° Solutions: 70 . 5 ° , 120 ° , 240 ° , 289 . 5 ° At this stage you could use the quadratic formula to solve the equation by letting y = cos x So, 6 y 2 + y - 1 = 0 Comparing with a y 2 + b y + c = 0 : a = 6 , b = 1 , c = - 1 y = - 1 ± 1 2 - 4 ( 6 ) ( - 1 ) 2 ( 6 ) y = - 1 ± 25 12 y = 1 3 or y = - 1 2 So, cos x = 1 3 or cos x = - 1 2 now continue as before.

Solve the equation 2 cos 2 x - sin 2 x - 2 sin x - 1 = 0 for 0 ∘ ⩽ x ⩽ 360 ∘ .

Given that, 2 cos 2 x - sin 2 x - 2 sin x - 1 = 0 , 0 ⩽ x ⩽ 360 ° ∴ 2 - 2 sin 2 x - sin 2 x - 2 sin x - 1 = 0 ∴ - 3 sin 2 x - 2 sin x + 1 = 0 ∴ 3 sin 2 x + 3 sin x - sin x - 1 = 0 ∴ 3 sin x ( sin x + 1 ) - ( sin x + 1 ) = 0 ∴ ( 3 sin x - 1 ) ( sin x + 1 ) = 0 ∴ 3 sin x - 1 = 0 or sin x + 1 = 0 ∴ sin x = 1 3 or sin x = - 1 Let, ∴ sin x = 1 3 ∴ x = sin - 1 1 3 ⇒ x = 17 . 457 ° o r 162 . 54 ° Also, ⇒ sin x = sin ( 270 ) = - 1 ⇒ x = 270 °

Solve the equation 1 + tan x cos x = 2 cos 2 x for 0 ∘ ⩽ x ⩽ 360 ∘ .

Given that, 1 + tan x cos x = 2 cos 2 x , 0 ⩽ x ≤ 360 ° ⇒ 1 + sin x cos x × cos x = 2 cos 2 x ⇒ 1 + sin x = 2 1 - sin 2 x ∴ 1 + sin x = 2 - 2 sin 2 x ∴ 2 sin 2 x + sin x - 1 = 0 ⇒ 2 sin x ( sin x + 1 ) - ( sin x + 1 ) = 0 ∴ 2 sin x - 1 ( sin x + 1 ) = 0 ∴ sin x + 1 = 0 or 2 sin x - 1 = 0 ⇒ sin x = - 1 o r sin x = 1 2 ⇒ sin x = - 1 ⇒ x = - 270 ° Also, sin x = 1 2 ⇒ x = 30 ° , 150 °

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>EXERCISE 5F

Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 9: EXERCISE 5F

Author:Sue Pemberton

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 9: EXERCISE 5F

Attempt the practice questions on Chapter 5: Trigonometry, Exercise 9: EXERCISE 5F with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book solutions are prepared by Experienced Embibe Experts.