Question Number 107978 by anonymous last updated on 13/Aug/20
![on the interval of [0,π] solve tan^2 x+3secx= −3](Q107978.png)
$${on}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\pi\right]\:{solve} \\ $$$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{3}{secx}=\:−\mathrm{3} \\ $$$$ \\ $$
Answered by bemath last updated on 14/Aug/20
$$\:\:\:\frac{\bullet\mathscr{B}{emath}\bullet}{\mathscr{C}{ooll}} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} {x}+\mathrm{3sec}\:{x}+\mathrm{3}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} {x}+\mathrm{3sec}\:{x}+\mathrm{2}=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{sec}\:{x}=−\mathrm{1}\Rightarrow\mathrm{cos}\:{x}=−\mathrm{1}}\\{\mathrm{sec}\:{x}=−\mathrm{2}\Rightarrow\mathrm{cos}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$
Commented by anonymous last updated on 13/Aug/20
$${Thank}\:{alot}\:{sif} \\ $$
Commented by aleks041103 last updated on 13/Aug/20
$${Should}\:{be}\:{cos}\:{x}=−\mathrm{1} \\ $$
Commented by bemath last updated on 14/Aug/20
$${yes}....{cooll} \\ $$