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General Solution of Trigonometric Equation

The general s...

Question

The general solution of x satisfying the equation $t a n 3 x - 1 = t a n 2 x (1 + t a n 3 x)$ , is

n π + \frac{π}{2}; n ε Z

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n π + \frac{3 π}{4}; n ε Z

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n π + \frac{π}{4}; n ε Z

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Non-existent

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Solution

The correct option is C $n π + \frac{π}{4}; n ε Z$
$tan 3 x - 1 = tan 2 x (1 + tan 3 x) \Rightarrow \frac{tan 3 x - 1}{1 + tan 3 x} = tan 2 x \Rightarrow tan (3 x - \frac{π}{4}) = tan 2 x \Rightarrow 3 x - \frac{π}{4} = 2 x + n π \Rightarrow x = \frac{π}{4} + n π (n \in Z)$
so C is the correct option

Suggest Corrections

Similar questions

\frac{tan 3 x - tan 1 x}{1 + tan 3 x tan 1 x} = 1

x = \frac{n π}{2} + \frac{π}{8}, n ϵ I

tan α = 1

α = n π + \frac{π}{4}, n ϵ I .

\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1 i s x = n π + \frac{π}{4}

tan x

\frac{π}{2}

\frac{tan 3 x - tan 2 x}{1 + tan 2 x tan 3 x} = 1

x = n π + \frac{π}{4}, \forall n \in I

tan x

x = n π + \frac{π}{2}, n \in I

\frac{t a n 3 x - t a n 2 x}{1 + t a n 3 x t a n 2 x} = 1 \Rightarrow x = n π + \frac{π}{4}, n \in l

t a n x

x = n π + \frac{π}{2}, n \in l

θ

{tan}^{2} θ + sec 2 θ = 1

n \in Z

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