What is the general solution of trigonometric equation ${sec}^{2} θ + 1 - 3 tan θ = 0$ $sec^2theta+1-3tantheta=0$ ?

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Answer 1 · 2016-10-06T14:35:38

$θ = n π + \frac{π}{4}$ $theta=npi+pi/4$

or $θ = n π + {tan}^{- 1} 2$ $theta=npi+tan^(-1)2$

As ${sec}^{2} θ = 1 + {tan}^{2} θ$ $sec^2theta=1+tan^2theta$ ,

${sec}^{2} θ + 1 - 3 tan θ = 0$ $sec^2theta+1-3tantheta=0$ can be written as

$1 + {tan}^{2} θ + 1 - 3 tan θ = 0$ $1+tan^2theta+1-3tantheta=0$

or ${tan}^{2} θ - 3 tan θ + 2 = 0$ $tan^2theta-3tantheta+2=0$

or ${tan}^{2} θ - 2 tan θ - tan θ + 2 = 0$ $tan^2theta-2tantheta-tantheta+2=0$

or $tan θ (tan θ - 2) - 1 (tan θ - 2) = 0$ $tantheta(tantheta-2)-1(tantheta-2)=0$

or $(tan θ - 1) (tan θ - 2) = 0$ $(tantheta-1)(tantheta-2)=0$

i.e. either $tan θ - 1 = 0$ $tantheta-1=0$ or $tan θ - 2 = 0$ $tantheta-2=0$

hence, either $tan θ = 1$ $tantheta=1$ or $tan θ = 2$ $tantheta=2$

i.e. $θ = n π + \frac{π}{4}$ $theta=npi+pi/4$

or $θ = n π + {tan}^{- 1} 2$ $theta=npi+tan^(-1)2$ i.e. $θ = n π + {tan}^{- 1} 2$ $theta=npi+tan^(-1)2$ (in radians using calculator).

What is the general solution of trigonometric equation sec^2theta+1-3tantheta=0sec2θ+1−3tanθ=0sec^2theta+1-3tantheta=0?