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

  1. Chapter 3 Class 11 Trigonometric Functions (Term 2)
  2. Serial order wise

Transcript

Ex 3.4, 5 Find the general solution of the equation cos 4x = cos 2x cos 4x = cos 2x cos 4x โ€“ cos 2x = 0 โ€“2 sin ((4๐‘ฅ + 2๐‘ฅ)/2) sin ((4๐‘ฅ โˆ’ 2๐‘ฅ)/2) = 0 โ€“2sin (6๐‘ฅ/2) sin (2๐‘ฅ/2) = 0 โ€“2 sin 3x sin x = 0 We know that cos x โ€“ cos y = โˆ’2sin (๐‘ฅ + ๐‘ฆ)/2 sin (๐‘ฅ โˆ’ ๐‘ฆ)/2 Replacing x with 4x and y with 2x sin 3x sin x = 0/(โˆ’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 = nฯ€ ยฑ (โ€“1)n 3y where n โˆˆ Z Putting y = 0 3x = nฯ€ ยฑ (โ€“1)n 0 3x = nฯ€ x = ๐‘›๐œ‹/3 where n โˆˆ 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 = nฯ€ ยฑ (โˆ’1)n y where n โˆˆ Z Putting y = 0 x = nฯ€ ยฑ (โˆ’1)n 0 x = nฯ€ where n โˆˆ Z Therefore, General Solution are For sin 3x = 0, x = ๐’๐…/๐Ÿ‘ Or For sin x = 0 , x = nฯ€ where n โˆˆ Z

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.