Đáp án:
Giải thích các bước giải:
`tan^2 x+ cot^2 x=2`
`ĐK: x \ne k\pi, x \ne \frac{k\pi}{2}`
`⇔ \frac{sin^2 x}{cos^2 x}+\frac{cos^2 x}{sin^2 x}=2`
`⇔ \frac{sin^4 x+cos^4 x}{cos^2 x. sin^2 x}=2`
`⇔ \frac{(sin^2 x+cos^2 x)^2-2sin^2 x.cos^2 x}{cos^2 x. sin^2 x}=2`
`⇔ 1-2sin^2 x-cos^2 x=2cos^2 x.sin^2 x`
`⇔ 1=4sin^2 x. cos^2 x`
`⇔ 1=2sin x.cos x. 2sin x. cos x`
`⇔ 1=sin^2 2x`
`⇔ sin 2x = ±1=sin \frac{\pi}{2}=sin \frac{-\pi}{2}`
` ⇔` \(\left[ \begin{array}{l}2x=(4k+1).\dfrac{\pi}{2}\ (k \in \mathbb{Z})\\2x=(4k-1).\dfrac{\pi}{2}\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=(4k+1).\dfrac{\pi}{4}\\x=(4k-1).\dfrac{\pi}{4}\end{array} \right.\) `(k \in \mathbb{Z})`
