Finding General Solutions

Chapter 3 Class 11 Trigonometric Functions
Concept wise

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### Transcript

Question 5 Find the general solution of the equation cos 4x = cos 2x cos 4x = cos 2x cos 4x β cos 2x = 0 β2 sin ((4π₯ + 2π₯)/2) sin ((4π₯ β 2π₯)/2) = 0 β2sin (6π₯/2) sin (2π₯/2) = 0 β2 sin 3x sin x = 0 We know that cos x β cos y = β2sin (π₯ + π¦)/2 sin (π₯ β π¦)/2 Replacing x with 4x and y with 2x sin 3x sin x = 0/(β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± (β1)n 3y where n β Z Putting y = 0 3x = nΟ Β± (β1)n 0 3x = nΟ x = ππ/3 where n β 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± (β1)n y where n β Z Putting y = 0 x = nΟ Β± (β1)n 0 x = nΟ where n β Z Therefore, General Solution are For sin 3x = 0, x = ππ/π Or For sin x = 0 , x = nΟ where n β Z