Question

giải phương trình

\(\frac{sin^22x-4\sin^2x}{sin^22x+4\sin^2x}+1=2\tan^2x\)

Answer 1

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4sin^2x}+1=2tan^2x\)

\(\Leftrightarrow\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x\left(cos^2x+1\right)}+1=\frac{2sin^2x}{cos^2x}\)

\(\Leftrightarrow\frac{cos^2x}{cos^2x+1}=\frac{1-cos^2x}{cos^2x}\)

Đặt \(cos^2x=t\Rightarrow0< t< 1\)

\(\Rightarrow\frac{t}{t+1}=\frac{1-t}{t}\Leftrightarrow t^2=1-t^2\Leftrightarrow t^2=\frac{1}{2}\)

\(\Leftrightarrow t=\frac{\sqrt{2}}{2}\Leftrightarrow cos^2x=\frac{\sqrt{2}}{2}\)