Ex 3.4, 6 - Find general solution of cos 3x + cos x - cos 2x = 0

Ex 3.4, 6 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.4, 6 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Ex 3.4, 6 - Chapter 3 Class 11 Trigonometric Functions - Part 4

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

Transcript

Ex 3.4, 6 Find the general solution of the equation cos 3x + cos x – cos 2x = 0 cos 3x + cos x – cos 2x = 0 (cos 3x + cos x) – cos 2x = 0 2 cos ((3π‘₯ + π‘₯)/2) . cos ((3π‘₯ βˆ’ π‘₯)/2) –cos 2x = 0 2 cos (4π‘₯/2) . cos (2π‘₯/2) βˆ’cos 2x = 0 2 cos 2x . cos x – cos 2x = 0 cos 2x (2cos x – 1) = 0 We know that cos x + cos y = 2 cos ((π‘₯ + 𝑦)/2) cos ((π‘₯ βˆ’ 𝑦)/2) Replacing x by 3x & y by x Hence We find general solutions of both separately General solution for cos 2x = 0 Given cos 2x = 0 Thus, general solution is 2x = (2n + 1) πœ‹/2 x = (2n + 1) 𝝅/πŸ’ where n ∈ Z (2cos x – 1) = 0 2cos x = 1 cos x = 1/2 General solution for cos x = 𝟏/𝟐 Let cos x = cos y Given cos x = 1/2 From (3) and (4) cos y = 1/2 cos y = cos (πœ‹/3) y = πœ‹/3 Rough We know that cos 60Β° = 1/2 So, 60Β° = 60 Γ— πœ‹/180 = πœ‹/3 General solution for cos x = cos y is x = 2nΟ€ Β± y where n ∈ Z Putting y = πœ‹/3 x = 2nΟ€ Β± πœ‹/3 where n ∈ Z Hence General Solution is For cos 2x = 0, x = (2n + 1) 𝝅/πŸ’ Or For cos x = 1/2 , x = 2nΟ€ Β± 𝝅/πŸ‘ 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 10 years. He provides courses for Maths and Science at Teachoo.