Question-203625

Question Number 203625 by professorleiciano last updated on 23/Jan/24

Answered by esmaeil last updated on 23/Jan/24

1+(((sin^2 x+cos^2 x)^2 −2sin^2 xcos^2 x)/(sin^2 xcos^2 x))=3→ 1=sin^2 2x→sin2x=±1=sin(±(π/2)) 2x=kπ±(π/2)→ { ((2x=2kπ+(π/2))),((2x=2kπ+π−(π/2))) :} { ((2x=2kπ−(π/2))),((2x=2kπ+((3π)/2))) :} x=kπ±(π/4) ,x=kπ+((3π)/4)

$$\mathrm{1}+\frac{\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}{{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}=\mathrm{3}\rightarrow \\ $$$$\mathrm{1}={sin}^{\mathrm{2}} \mathrm{2}{x}\rightarrow{sin}\mathrm{2}{x}=\pm\mathrm{1}={sin}\left(\pm\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{2}{x}={k}\pi\pm\frac{\pi}{\mathrm{2}}\rightarrow\begin{cases}{\mathrm{2}{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}}\\{\mathrm{2}{x}=\mathrm{2}{k}\pi+\pi−\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2}{x}=\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}}}\\{\mathrm{2}{x}=\mathrm{2}{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{2}}}\end{cases} \\ $$$${x}={k}\pi\pm\frac{\pi}{\mathrm{4}}\:\:\:,{x}={k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$ \\ $$

Commented by professorleiciano last updated on 24/Jan/24

Sua resposta está totalmente errada.

Answered by professorleiciano last updated on 23/Jan/24