Ex 3.4, 5 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Sept. 3, 2021 by Teachoo
Last updated at Sept. 3, 2021 by Teachoo
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 โ2 sin ((4๐ฅ + 2๐ฅ)/2) sin ((4๐ฅ โ 2๐ฅ)/2) = 0 โ2sin (6๐ฅ/2) sin (2๐ฅ/2) = 0 โ2 sin 3x sin x = 0 We know that cos x โ cos y = โ2sin (๐ฅ + ๐ฆ)/2 sin (๐ฅ โ ๐ฆ)/2 Replacing x with 4x and y with 2x sin 3x sin x = 0/(โ2) 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 (โ1)n 3y where n โ Z Putting y = 0 3x = nฯ ยฑ (โ1)n 0 3x = nฯ x = ๐๐/3 where n โ 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 (โ1)n y where n โ Z Putting y = 0 x = nฯ ยฑ (โ1)n 0 x = nฯ where n โ Z Therefore, General Solution are For sin 3x = 0, x = ๐๐ /๐ Or For sin x = 0 , x = nฯ where n โ Z
Ex 3.4
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Ex 3.4
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