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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.