The solution for x which satisfy the equation tan 3x-tan 2x1-tan 3xtan 2x-1 is

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Integration Using Substitution

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Question

The solution for $x$ which satisfy the equation $\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1$ is

x = n π + \frac{π}{4}, n \in l

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x = 2 n π + \frac{π}{4}, n \in l

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Both

(a)

and

(b)

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No such

x

exists

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\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

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

{n π + \frac{π}{4}, n \in l}

(1 + t a n x + {tan}^{2} x)

(1 - cot x + {cot}^{2} x)

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e^{sin x} - e^{- s ˙ i n x} \neq 4

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The general solution to $s i n^{10} x + c o s^{10} x = \frac{29}{16} c o s^{4} 2 x$ is

\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

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cos x = \frac{- 1}{\sqrt{2}}

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