Math, asked by goravsharma0245, 4 months ago

If x belongs to [0,2π] then number of values of X satisfying the equation tan8x = (cos x - sin x)/(cos x + sin x) is​

Answers

Answered by amitnrw
7

Given : x ∈  [0,2π]  , tan8x = (cos x - sin x)/(cos x + sin x)

To Find  :  number of values of x satisfying the equation

Solution:

Tan8x = (cosx - Sinx)/(Cosx + Sinx)

using Sinx = Cosx.Tanx

=>Tan8x = Cosx(1 - Tanx)/Cosx(1 + Tanx)

=> Tan8x = (1 -Tanx)/(1 + Tanx)

=> Tan8x = (1 -Tanx)/(1 + 1.Tanx)

1 = Tan(nπ+ π/4)   n ∈ Z

=>   Tan8x = (Tan(nπ+π/4) -Tanx)/(1 + Tan(nπ+π/4).Tanx)

Tan(A - B)  = (Tan A - TanB)/(1 + TanA.TanB)

=> Tan8x =  Tan(nπ+π/4  - x)

8x = nπ + π/4  - x

=> 9x =  nπ + π/4  

=> x =   (4nπ  + π)/36

x ∈  [0,2π]

0 ≤  (4nπ  + π)/36 ≤ 2π

=> 0 ≤  (4nπ  + π)  ≤ 72π

=> 0 ≤   4nπ  + π   ≤ 72π

=>  0 ≤   4nπ  + π   ≤ 72π

=> -π  ≤   4nπ ≤ 71π

=> -1  ≤   4n  ≤ 71

=>  -1/4  ≤    n  ≤ 71/4

=>  n = 0 to  17

18 possible values

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