Find the general solution of the following equation:
$tan θ + cot 2 θ = 0$

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Solution

$tan θ + cot 2 θ = 0 (Given)$
$\Rightarrow tan θ = - cot 2 θ$
$\Rightarrow tan θ = tan (\frac{π}{2} + 2 θ) (∵ tan (\frac{π}{2} + x) = - cot x)$
$\Rightarrow θ = n π + (\frac{π}{2} + 2 θ); n \in Z$
$\Rightarrow - θ = n π + \frac{π}{2}$
$\Rightarrow θ = - (n π + \frac{π}{2})$

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Find the general solution of the following equation:tanθ+cot2θ=0

tanθ+cot2θ=0(Given)⇒tanθ=−cot2θ⇒tanθ=tan(π2+2θ)(∵tan(π2+x)=−cotx)⇒θ=nπ+(π2+2θ);n∈Z⇒−θ=nπ+π2⇒θ=−(nπ+π2)

Find the general solution of the following equation:
$tan θ + cot 2 θ = 0$

$tan θ + cot 2 θ = 0 (Given)$
$\Rightarrow tan θ = - cot 2 θ$
$\Rightarrow tan θ = tan (\frac{π}{2} + 2 θ) (∵ tan (\frac{π}{2} + x) = - cot x)$
$\Rightarrow θ = n π + (\frac{π}{2} + 2 θ); n \in Z$
$\Rightarrow - θ = n π + \frac{π}{2}$
$\Rightarrow θ = - (n π + \frac{π}{2})$