# Solve: tan5X = cot 2X Please try solving asap. Thank you:) You should move `cot 2x`  to the left side such that:

`tan 5x - cot 2x = 0 `

You may write `cot 2x = tan(pi/2 - 2x)`  such that:

`tan 5x - tan(pi/2 - 2x) = 0 `

You need to convert the difference into a product such that:

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You should move `cot 2x`  to the left side such that:

`tan 5x - cot 2x = 0 `

You may write `cot 2x = tan(pi/2 - 2x)`  such that:

`tan 5x - tan(pi/2 - 2x) = 0 `

You need to convert the difference into a product such that:

`(sin(5x - pi/2 + 2x))/(cos 5x cos (pi/2 - 2x)) = 0 `

The fraction is equal to zero if `sin(7x - pi/2) = 0`  and `(cos 5x cos (pi/2 - 2x)) != 0`  such that:

`sin(7x - pi/2) = 0 => 7x - pi/2 = (-1)^n arcsin 0 + npi`

`7x - pi/2 = npi => 7x = npi + pi/2 => x = npi/7 + pi/14`

Hence, evaluating the general solution to the given equation yields `x = npi/7 + pi/14.`

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