Đáp án:
$x = k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin^{2018}x + \cos^{2017}x = 1$
$\Leftrightarrow\sin^{2018}x + \cos^{2017}x =\sin^2x + \cos^2x$
$\Leftrightarrow \sin^{2018}x - \sin^2x = \cos^2x - \cos^{2017}x$
$\Leftrightarrow \sin^2x(\sin^{2016}x - 1) = \cos^2x(1 - \cos^{2015}x)$
Ta có:
$VT = \sin^2x(\sin^{2016}x - 1)$
$0\leq \sin^2x \leq 1$
$0 \leq \sin^{2016}x \leq 1$
$\Rightarrow \sin^{2016}x - 1 \leq 0$
$\Rightarrow \sin^2x(\sin^{2016}x - 1) \leq 0$
$\Rightarrow VT \leq 0$
$VP = \cos^2x(1 - \cos^{2015}x)$
$0 \leq \cos^2x \leq 1$
$- 1 \leq \cos^{2015}x \leq 1$
$\Rightarrow 1 -\cos^{2015}x \geq 0$
$\Rightarrow \cos^2x(1 - \cos^{2015}x) \geq 0$
$\Rightarrow VP \geq 0$
$\Rightarrow VT = VP$
$\Leftrightarrow \begin{cases}\sin^2x(\sin^{2016}x - 1) = 0\\\cos^2x(1 - \cos^{2015}x) = 0\end{cases}$
$\Leftrightarrow \left[\begin{array}{l}\begin{cases}\sin x = 0\\\cos x = 1\end{cases}\\\begin{cases}\sin x = 1\\\cos x = 0\end{cases}\end{array}\right.$
$\Leftrightarrow x = k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$
