The number of solutions in [0,2π] of the equation cot 2x/cot x + cot x/cot 2x + 2 = 0 is
Answers
Given : cot 2x/cot x + cot x/cot 2x + 2 = 0 x ∈ [0,2π]
To find : Number of Solutions
Solution:
Cot2x/Cotx + Cotx/Cot2x + 2 = 0
=> Cos2x.Sinx/Sin2xCosx + Cosx.Sin2x/Sinx. Cos2x + 2 = 0
=> Cos2x.Sinx/2SinxCosxCosx + cosx2SinxCosx/SinxCos2x + 2 = 0
=> Cos2x/2Cos²x + 2Cos²x/Cos2x + 2 = 0
=> (2Cos²x - 1)/2Cos²x + (1 + Cos2x )/Co2x + 2 = 0
=> 1 - 1/2Cos²x + 1/Cos2x + 1 + 2 = 0
=> 1/Cos2x - 1/2Cos²x = - 4
=> 1/(2Cos²x - 1) - 1/2Cos²x = - 4
=> 2Cos²x - 2Cos²x + 1 = -8 (2Cos²x - 1) Cos²x
=> 1 = -16Cos⁴x + 8Cos²x
=> 16Cos⁴x - 8Cos²x + 1 = 0
=> (4Cos²x - 1)² = 0
=> 4Cos²x - 1 = 0
=> 4Cos²x = 1
=> Cos²x = 1/4
=> Cosx = ± 1/2
x = nπ ± π/3
π/3 , 2 π/3 , 3 π/3 , 5 π/3
4 Solutions
another way
cot 2x/cot x + cot x/cot 2x + 2 = 0 is
Let say cot 2x/cot x = a
=> a + 1/a + 2 = 0
=> a² + 2a + 1 = 0
=> (a + 1)² = 0
=> a = - 1
cot 2x/cot x = - 1
=> Cot2x = -Cotx
=> 1/tan2x = -1/tanx
=> tan2x = - Tanx
=> 2Tanx/(1 - Tan²x) = - Tanx
=> 2 = Tan²x - 1
=> Tan²x = 3
=> Tanx = ±√3
x = nπ ± π/3
π/3 , 2 π/3 , 3 π/3 , 5 π/3
4 Solutions
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