Question Number 195263 by dimentri last updated on 28/Jul/23
$$\:\:\:\:\:\:\begin{cases}{{y}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\mathrm{2}}\\{\mathrm{4sin}\:{x}−\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=?\: \\ $$
Answered by cortano12 last updated on 28/Jul/23
$$\:\:\:\:\:\:\begin{cases}{\:\cancel{\underline{\underbrace{ }}}}\end{cases} \\ $$
Answered by Frix last updated on 28/Jul/23
![{ ((y=((2−cos x)/(sin x)))),((y=((4sin x)/(1+2cos x)))) :} ((2−cos x)/(sin x))=((4sin x)/(1+2cos x)) cos^2 x +(3/2)cos x −1=0 cos x =(1/2) [∨cos x =−2] tan x =±((√(1−cos^2 x))/(cos x))=±(√3)](https://www.tinkutara.com/question/Q195272.png)
$$\begin{cases}{{y}=\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}\\{{y}=\frac{\mathrm{4sin}\:{x}}{\mathrm{1}+\mathrm{2cos}\:{x}}}\end{cases} \\ $$$$\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{4sin}\:{x}}{\mathrm{1}+\mathrm{2cos}\:{x}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:{x}\:−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\left[\vee\mathrm{cos}\:{x}\:=−\mathrm{2}\right] \\ $$$$\mathrm{tan}\:{x}\:=\pm\frac{\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}}}{\mathrm{cos}\:{x}}=\pm\sqrt{\mathrm{3}} \\ $$