on-the-interval-of-0-pi-solve-tan-2-x-3secx-3-

Question Number 107978 by anonymous last updated on 13/Aug/20

on the interval of [0,π] solve tan^2 x+3secx= −3

$${on}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\pi\right]\:{solve} \\ $$$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{3}{secx}=\:−\mathrm{3} \\ $$$$ \\ $$

Answered by bemath last updated on 14/Aug/20

((•Bemath•)/(Cooll)) ⇒sec^2 x+3sec x+3−1=0 ⇒sec^2 x+3sec x+2=0 { ((sec x=−1⇒cos x=−1)),((sec x=−2⇒cos x=−(1/2))) :}

$$\:\:\:\frac{\bullet\mathscr{B}{emath}\bullet}{\mathscr{C}{ooll}} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} {x}+\mathrm{3sec}\:{x}+\mathrm{3}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} {x}+\mathrm{3sec}\:{x}+\mathrm{2}=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{sec}\:{x}=−\mathrm{1}\Rightarrow\mathrm{cos}\:{x}=−\mathrm{1}}\\{\mathrm{sec}\:{x}=−\mathrm{2}\Rightarrow\mathrm{cos}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

Commented by anonymous last updated on 13/Aug/20