Question 5 - Principal and General Solutions - Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
Principal and General Solutions
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Principal and General Solutions
Last updated at April 16, 2024 by Teachoo
Question 5 Find the general solution of the equation cos 4x = cos 2x cos 4x = cos 2x cos 4x â cos 2x = 0 â2 sin ((4đĨ + 2đĨ)/2) sin ((4đĨ â 2đĨ)/2) = 0 â2sin (6đĨ/2) sin (2đĨ/2) = 0 â2 sin 3x sin x = 0 We know that cos x â cos y = â2sin (đĨ + đĻ)/2 sin (đĨ â đĻ)/2 Replacing x with 4x and y with 2x sin 3x sin x = 0/(â2) 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 = nĪ Âą (â1)n 3y where n â Z Putting y = 0 3x = nĪ Âą (â1)n 0 3x = nĪ x = đđ/3 where n â 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 = nĪ Âą (â1)n y where n â Z Putting y = 0 x = nĪ Âą (â1)n 0 x = nĪ where n â Z Therefore, General Solution are For sin 3x = 0, x = đđ /đ Or For sin x = 0 , x = nĪ where n â Z