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**Hint:**Trigonometric functions describe the relation between the sides and angles of a right triangle. ... The trigonometric functions include the following 6 functions: sine, cosine, tangent, cotangent, secant, and cosecant. For each of these functions, there is an inverse trigonometric function.

Always convert All the Trigonometric Function into the same function.

As to convert \[\tan x\] into \[\cot x\] the following ways are :

\[\cot x = \tan (\dfrac{\pi }{2} - y)\]

Formula for \[\tan \theta = \tan \alpha \] this is the solution.

\[ \Rightarrow \theta = n\pi + \alpha \]

**Complete step-by- step solution:**

Given \[\tan 5x = \cot 3x\]

Changing the \[cot\theta \]into \[tan\theta \] by writing \[cot\theta \]= \[tan\theta \]

Where \[\theta = 3x\]

\[\tan 5x = \tan (\dfrac{\pi }{2} - 3x)\]

For General formula of \[tan{\text{ }}x,\] when \[tan\theta = tan{\text{ }}\alpha \]

Then, \[\theta = n\pi {\text{ }} + \alpha \]

$\Rightarrow$\[5x = n\pi + \dfrac{\pi }{2} - 3x\]

Now on shifting \[\pi /2\] on right side we get

$\Rightarrow$\[5x + 3x = n\pi + \dfrac{\pi }{2}\]

$\Rightarrow$\[8x = n\pi + \dfrac{\pi }{2}\]

Now on taking L.CM for the above term where L.C.M = 2 we get

$\Rightarrow$\[8x = \dfrac{{2n\pi + \pi }}{2}\]

Transposing 2 to left side and multiply with 8 we get

$\Rightarrow$\[8x \times 2 = 2n\pi + \pi \]

$\Rightarrow$\[16x = 2n\pi + \pi \]

Transposing 16 on right side as denominator

$\Rightarrow$\[x = \dfrac{1}{{16}}(2\pi n + \pi )\]

Hence Required value of \[x = \dfrac{1}{{16}}(2n\pi + \pi )\]

Or \[x = \dfrac{\pi }{{16}}(2n + 1)\] [Taking \[\pi \] common]

**Note:**Explanation for General solution of \[\tan \theta \]\[ = \tan \alpha \] is given by \[\theta = n\pi + \alpha ,\,n \in z\].

As follows,

\[\tan \theta = \tan \alpha \]

We know \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]

$\Rightarrow$ \[\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\sin \alpha }}{{\cos \alpha }}\]

Taking \[\dfrac{{\sin \alpha }}{{\cos \alpha }}\] to L.H.S we get

$\Rightarrow$ \[\dfrac{{\sin \theta }}{{\cos \theta }} - \dfrac{{\sin \alpha }}{{\cos \alpha }} = 0\]

On taking taking LCM we get

$\Rightarrow$\[\dfrac{{\sin \theta \cos \alpha - \sin \alpha \cos \theta }}{{\cos \theta \cos \alpha }} = 0\]

$\Rightarrow$\[\therefore \cos \theta \cos \alpha \times 0 = 0\]

$\Rightarrow$\[\sin \theta \cos \alpha - \sin \alpha \cos \theta = 0\]

Identity: \[\sin \theta \cos \alpha - \sin \alpha \cos \theta \] ,\[\sin (\theta - \alpha )\]

\[\sin (\theta - \alpha ) = 0\]

\[ \Rightarrow \theta - \alpha = n\pi \], where \[n \in z\] i.e. \[(n = 0, \pm 1, \pm 2, \pm 3.....)\].

(since we know that \[\theta = n\pi ,n \in z\] is the general solution of the given equation \[\sin \theta = 0\]).

\[ \Rightarrow \theta = n\pi + \alpha ,\] where \[n \in z\] (i.e. \[n = 0,\,\, \pm 1,\,\,\, \pm 2,\,\, \pm 3).\]

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