Đáp án:
n. $x = \dfrac{\pi}{4} + k\pi$ $(k \in \Bbb Z)$
t. $\left[\begin{array}{l}x = k\dfrac{\pi}{4}\\x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\end{array}\right.$ $(k \in \Bbb Z)$
Giải thích các bước giải:
n. $tanx + cotx - 2 = 0$ $(*)$
$ĐK: sin2x \ne 0 \Leftrightarrow x \ne k\dfrac{\pi}{2}$
$(*) \Leftrightarrow \dfrac{sinx}{cosx} + \dfrac{cosx}{sinx} - 2 = 0$
$\Leftrightarrow \dfrac{sin^2x + cos^2x}{sinx.cosx} - 2 = 0$
$\Leftrightarrow 1 - 2sinx.cosx = 0$
$\Leftrightarrow 1 - sin2x = 0$
$\Leftrightarrow sin2x = 1$
$\Leftrightarrow 2x = \dfrac{\pi}{2} + k2\pi$
$\Leftrightarrow x = \dfrac{\pi}{4} + k\pi$ (thoả ĐK)
Vậy $x = \dfrac{\pi}{4} + k\pi$ $(k \in \Bbb Z)$
t. $4cos^5xsinx - 4sin^5xcosx = sin^24x$
$\Leftrightarrow 4sinxcosx(cos^4x - sin^4x) = sin^24x$
$\Leftrightarrow 2sin2x(cos^2x+sin^2x)(cos^2x - sin^2x) = sin^24x$
$\Leftrightarrow 2sin2x.cos2x = sin^24x$
$\Leftrightarrow sin4x = sin^24x$
$\Leftrightarrow sin4x(sin4x - 1) = 0$
$\Leftrightarrow \left[\begin{array}{l}sin4x = 0\\sin4x = 1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}4x = k\pi\\4x = \dfrac{\pi}{2} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{4}\\x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\end{array}\right.$ $(k \in \Bbb Z)$
Vậy $\left[\begin{array}{l}x = k\dfrac{\pi}{4}\\x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\end{array}\right.$ $(k \in \Bbb Z)$
