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Ex 3.4, 5 - Find general solution of cos 4x = cos 2x - Chapter 3

Ex 3.4, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.4, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Ex 3.4, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 4
Ex 3.4, 5 - Chapter 3 Class 11 Trigonometric Functions - Part 5

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Transcript

Ex 3.4, 5 Find the general solution of the equation cos 4x = cos 2x cos 4x = cos 2x cos 4x – cos 2x = 0 –2 sin ((4π‘₯ + 2π‘₯)/2) sin ((4π‘₯ βˆ’ 2π‘₯)/2) = 0 –2sin (6π‘₯/2) sin (2π‘₯/2) = 0 –2 sin 3x sin x = 0 We know that cos x – cos y = βˆ’2sin (π‘₯ + 𝑦)/2 sin (π‘₯ βˆ’ 𝑦)/2 Replacing x with 4x and y with 2x sin 3x sin x = 0/(βˆ’2) sin 3x sin x = 0 So, either sin 3x = 0 or sin x = 0 We solve sin 3x = 0 & sin x = 0 separately General solution for sin 3x = 0 Let sin x = sin y sin 3x = sin 3y Given sin 3x = 0 From (1) and (2) sin 3y = 0 sin 3y = sin (0) 3y = 0 y = 0 General solution for sin 3x = sin 3y is 3x = nΟ€ Β± (–1)n 3y where n ∈ Z Putting y = 0 3x = nΟ€ Β± (–1)n 0 3x = nΟ€ x = π‘›πœ‹/3 where n ∈ Z General solution for sin x = 0 Let sin x = sin y Given sin x = 0 From (1) and (2) sin y = 0 sin y = sin (0) y = 0 General solution for sin x = sin y is x = nΟ€ Β± (βˆ’1)n y where n ∈ Z Putting y = 0 x = nΟ€ Β± (βˆ’1)n 0 x = nΟ€ where n ∈ Z Therefore, General Solution are For sin 3x = 0, x = 𝒏𝝅/πŸ‘ Or For sin x = 0 , x = nΟ€ where n ∈ Z

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.