Tan²x + cot²x = 2 ( giải PT )

Đáp án:

\(S = \left\{\dfrac{\pi}{4} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\)

Giải thích các bước giải:

\(\begin{array}{l}
\quad \tan^2x + \cot^2x = 2\qquad (*)\\
ĐK: \begin{cases}\sin x \ne 0\\\cos x \ne 0\end{cases}\Leftrightarrow \sin2x \ne 0\Leftrightarrow  x \ne \dfrac{n\pi}{2}\\
(*) \Leftrightarrow \dfrac{\sin^4x + \cos^4x}{\sin^2x\cos^2x} = 2\\
\Leftrightarrow \sin^4x + \cos^4x - 2\sin^2x\cos^2x =0\\
\Leftrightarrow (\sin^2x + \cos^2x)^2 - 4\sin^2x\cos^2x =0\\
\Leftrightarrow 1 - \sin^22x = 0\\
\Leftrightarrow 1 - \dfrac{1 -\cos4x}{2} = 0\\
\Leftrightarrow \cos4x = -1\\
\Leftrightarrow 4x = \pi + k2\pi\\
\Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{4} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)