The number of solutions of the equation tan2x · tanx = 1, x ∈ [0, 2π] is
Answers
Given : tan2x · tanx = 1, x ∈ [0, 2π]
To find : number of solutions of the equation
Solution:
tan2x · tanx = 1,
tan2x = 2Tanx/(1 - Tan²x)
=> ( 2Tanx/(1 - Tan²x)) ·Tanx = 1,
=> 2Tan²x = 1 - Tan²x
=> 3Tan²x = 1
=> Tan²x = 1/3
=> Tanx = ±1/√3
=> x = nπ ± π/6
x ∈ [0, 2π]
=> x = π/6 , 5π/6 , 7π/6 , 11π/6
Hence 4 Solutions
4 number of solutions of the equation tan2x · tanx = 1,
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Step-by-step explanation:
Given : tan2x · tanx = 1, x ∈ [0, 2π]
To find : number of solutions of the equation
Solution:
tan2x · tanx = 1,
tan2x = 2Tanx/(1 - Tan²x)
=> ( 2Tanx/(1 - Tan²x)) ·Tanx = 1,
=> 2Tan²x = 1 - Tan²x
=> 3Tan²x = 1
=> Tan²x = 1/3
=> Tanx = ±1/√3
=> x = nπ ± π/6
x ∈ [0, 2π]
=> x = π/6 , 5π/6 , 7π/6 , 11π/6
Hence 4 Solutions
4 number of solutions of the equation tan2x · tanx = 1,