Find the general solution of equation cot⁡x+tan⁡x=2 My approach: cot⁡x+tan⁡x=(1+tan2x)tan⁡x=2 sec2x+tan⁡x=2 1(sin⁡xcos⁡x)=2 sin2x+cos2x=2sin⁡xcos⁡x (sin⁡x−cos⁡x)2=0 sin⁡x=cos⁡x So, x=nπ+(π4). This was my answer. But the answer given is x=2nπ±π3

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Find the general solution of equation cot⁡x+tan⁡x=2  My approach:  cot⁡x+tan⁡x=(1+tan2x)tan⁡x=2    sec2x+tan⁡x=2    1(sin⁡xcos⁡x)=2    sin2x+cos2x=2sin⁡xcos⁡x    (sin⁡x−cos⁡x)2=0    sin⁡x=cos⁡x   So, x=nπ+(π4). This was my answer. But the answer given is x=2nπ±π3

Ben Owens · Accepted Answer

You know that cot⁡(x)=1tan⁡(x). Take random letter, let a=tan⁡(x)   Then it is  1a+a=2   1+a2=2a   a2−2a+1=0   (a−1)2=0   a−1=0   a=1   tan⁡(x)=1

servidopolisxv · Accepted Answer

wrong reduction to algebraic form.   tan⁡x≡u⇒u+1u=2⇒u=1   u=1⇒x=nπ+π4

Vasquez · Accepted Answer

Note that it is enough to consider x&amp;gt;0. By AM-Gm inequality cot⁡x+tan⁡x2≥cot⁡xtan⁡x=1 for all x⧸∈{nπ2|n∈Z}Z and equality occur when cot⁡x=tan⁡x which implies x∈{nπ+π4|n∈Z}

Find the general solution of equation \cot x+\tan x=2 My approach: \cot

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