1. DK: $\sin x \neq 0$ hay $x \neq k\pi$.

Nhân chéo lên ta có

$1 + \sin(2x) + \cos(2x) = \sqrt{2} \sin x .2\sin x \cos x (1 + \dfrac{\cos^2x}{\sin^2x})$

$<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} .\sin^2x \cos x(1 + \dfrac{\cos^2x}{\sin^2x})$

$<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} .\sin^2x \cos x + 2\sqrt{2} \cos^3x$

$<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} .\cos x(\sin^2x + \cos^2x)$

$<-> 1 + \sin(2x) + \cos(2x) = 2\sqrt{2} \cos x$

$<-> 1 + 2\sin x \cos x + 2\cos^2x - 1 = 2\sqrt{2} \cos x$

$<-> \sin x \cos x + \cos^2x = \sqrt{2} \cos x$

Vậy $\cos x = 0$ nên $x = \dfrac{\pi}{2} + k\pi$ hoặc

$\sin x + \cos x = \sqrt{2}$

$<-> \sqrt{2} (\dfrac{1}{\sqrt{2}} \sin x + \dfrac{1}{\sqrt{2}} \cos x ) = \sqrt{2}$

$<-> \sin x \cos \dfrac{\pi}{4} + \sin \dfrac{\pi}{4} \cos x = 1$

$<-> \sin(x + \dfrac{\pi}{4}) = 1$

Vậy $x + \dfrac{\pi}{4} = \dfrac{\pi}{2} + 2k\pi$ hay $x = \dfrac{\pi}{4} + 2k\pi$.

Vậy nghiệm của ptrinh là $x = \dfrac{\pi}{2} + k\pi$ hoặc $x = \dfrac{\pi}{4} + 2k\pi$.

2. ĐK: $\sin x \neq \dfrac{\sqrt{2}}{2}$ hay $x \neq \dfrac{\pi}{4} + 2k\pi$ hoặc $x \neq \dfrac{3\pi}{4} + 2k\pi$.

Ptrinh tương đương vs

$\cos^6x + \sin^6x - \sin x \cos x = 0$

$<-> (\cos^2x + \sin^2x) (\cos^4x - \sin^2x \cos^2x + \sin^4x) - \sin x \cos x = 0$

$<-> \cos^4x + \sin^4x - \sin^2x \cos^2x - \sin x \cos x = 0$

$<-> (\cos^2x + \sin^2x) - 3\sin^2x \cos^2x - \sin x \cos x = 0$

$<-> 1 - 3\sin^2x \cos^2x - \sin x \cos x = 0$

$<-> 1 - \dfrac{3}{4} \sin^2(2x) - \dfrac{1}{2} \sin(2x) = 0$

Khi đó ta có

$\sin(2x) = \dfrac{-1 + \sqrt{13}}{3}, \sin(2x) = \dfrac{-1 - \sqrt{13}}{3}$ (loại do $\dfrac{-1 - \sqrt{13}}{3} < -1$)

Vậy

$2x = \arcsin(\dfrac{-1 + \sqrt{13}}{3}) + 2k\pi, 2x = \pi - \arcsin(\dfrac{-1 + \sqrt{13}}{3}) + 2k\pi$

Do đó

$x = \dfrac{\arcsin(\dfrac{-1 + \sqrt{13}}{3})}{2} + k\pi, x = \dfrac{\pi}{2} - \dfrac{\arcsin(\dfrac{-1 + \sqrt{13}}{3})}{2} + k\pi$.