Ta có:
$\left\{\begin{array}{l}\sin^2(3x-\dfrac{\pi}{4})\ge0\\\cos^2(x+\dfrac{\pi}{3})\ge0\end{array}\right.\Rightarrow \sin^2(3x-\dfrac{\pi}{4})+\cos^2(x+\dfrac{\pi}{3})\ge0$
Dấu bằng xảy ra khi:
      $\left\{\begin{array}{l}\sin^2(3x-\dfrac{\pi}{4})=0\\\cos^2(x+\dfrac{\pi}{3})=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sin(3x-\dfrac{\pi}{4})=0\\\cos(x+\dfrac{\pi}{3})=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}3x-\dfrac{\pi}{4}=k\pi&,k\in\mathbb{Z}\\x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+l\pi&,l\in\mathbb{Z}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=\dfrac{\pi}{12}+k\dfrac{\pi}{3}&,k\in\mathbb{Z}\\x=\dfrac{\pi}{6}+l\pi&,l\in\mathbb{Z}\end{array}\right.$
Suy ra phương trình vô nghiệm.

Ta có:
$\left\{\begin{array}{l}\sin^62x+1\ge 1\\\sin^2x\le1\end{array}\right.\Rightarrow \sin^62x+1\ge\sin^2x$
Dấu bằng xảy ra khi:
      $\left\{\begin{array}{l}\sin2x=0\\\sin^2x=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\sin2x=0\\\cos x=0\end{array}\right.$
$\Leftrightarrow \cos x=0$
$\Leftrightarrow x=\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}$

Ta có:
$\left\{\begin{array}{l}\cos\dfrac{3x}{4}\le 1\\\cos 2x\le1\end{array}\right.\Rightarrow \cos\dfrac{3x}{4}+\cos2x\le 2$
Dấu bằng xảy ra khi:
      $\left\{\begin{array}{l}\cos\dfrac{3x}{4}=1\\\cos 2x=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\dfrac{3x}{4}=k\pi&,k\in\mathbb{Z}\\2x=l\pi&,l\in\mathbb{Z}\end{array}\right.$
$\Leftrightarrow x=k4\pi,k\in\mathbb{Z}$

Phương trình đã cho tương đương với:
      $(\sin x-\cos x)(\sin^2x+\sin x\cos x+\cos^2x)=3\sin x\cos x-1$
$\Leftrightarrow (\sin x-\cos x)(1+\sin x\cos x)=3\sin x\cos x-1$
Đặt: $t=\sin x-\cos x=\sqrt2\sin(x-\dfrac{\pi}{4}),|t|\le\sqrt2$, khi đó: $\sin x\cos x=\dfrac{1-t^2}{2}$
Phương trình trở thành:
      $t(1+\dfrac{1-t^2}{2})=3\dfrac{1-t^2}{2}-1$
$\Leftrightarrow t^3-3t^2-3t+1=0$
$\Leftrightarrow \left[\begin{array}{l}t=-1\\t=2-\sqrt3\end{array}\right.$, vì $|t|\le\sqrt2$
Khi đó:
      $\left[\begin{array}{l}\sin(x-\dfrac{\pi}{4})=\dfrac{-1}{\sqrt2}\\\sin(x-\dfrac{\pi}{4})=\dfrac{2-\sqrt3}{\sqrt2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=k2\pi\\x=\dfrac{3\pi}{2}+k2\pi\\x=\arcsin\dfrac{2-\sqrt3}{\sqrt2}+\dfrac{\pi}{4}+k2\pi\\x=\pi-\arcsin\dfrac{2-\sqrt3}{\sqrt2}+\dfrac{\pi}{4}+k2\pi\end{array}\right.,k\in\mathbb{Z}$

Đặt: $t=\sin x-\cos x=\sqrt2\sin(x-\dfrac{\pi}{4}),|t|\le\sqrt2$, khi đó: $\sin x\cos x=\dfrac{1-t^2}{2}$
Phương trình trở thành:
      $t=2\sqrt6\dfrac{1-t^2}{2}$
$\Leftrightarrow \sqrt6t^2+t-\sqrt6=0$
$\Leftrightarrow \left[\begin{array}{l}t=\dfrac{2}{\sqrt6}\\t=\dfrac{-\sqrt6}{2}\end{array}\right.$
Khi đó:
      $\left[\begin{array}{l}\sin(x-\dfrac{\pi}{4})=\dfrac{1}{\sqrt3}\\\sin(x-\dfrac{\pi}{4})=\dfrac{-\sqrt3}{2}\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\arcsin\dfrac{1}{\sqrt3}+\dfrac{\pi}{4}+k2\pi\\x=\pi-\arcsin\dfrac{1}{\sqrt3}+\dfrac{\pi}{4}+k2\pi\\x=\dfrac{-\pi}{12}+k2\pi\\x=\dfrac{19\pi}{12}+k2\pi\end{array}\right.,k\in\mathbb{Z}$

Đặt: $t=\sin x-\cos x=\sqrt2\sin(x-\dfrac{\pi}{4}),|t|\le\sqrt2$, khi đó: $\sin x\cos x=\dfrac{1-t^2}{2}$
Phương trình trở thành:
      $t=-1-4\dfrac{1-t^2}{2}$
$\Leftrightarrow 2t^2-t-3=0$
$\Leftrightarrow t=-1$, vì $|t|\le\sqrt2$
Khi đó:
      $\sin(x-\dfrac{\pi}{4})=\dfrac{-1}{\sqrt2}$
$\Leftrightarrow \left[\begin{array}{l}x=k2\pi\\x=\dfrac{3\pi}{2}+k2\pi\end{array}\right.,k\in\mathbb{Z}$

Phương trình đã cho tương đương với:
      $(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=\dfrac{\sqrt2}{2}$
$\Leftrightarrow (\sin x+\cos x)(1-\sin x\cos x)=\dfrac{\sqrt2}{2}$
Đặt: $t=\sin x+\cos x=\sqrt2\sin(x+\dfrac{\pi}{4}),|t|\le\sqrt2$, khi đó: $\sin x\cos x=\dfrac{t^2-1}{2}$
Phương trình trở thành:
      $t(1-\dfrac{t^2-1}{2})=\dfrac{\sqrt2}{2}$
$\Leftrightarrow t^3-3t+\sqrt2=0$
$\Leftrightarrow \left[\begin{array}{l}t=\sqrt2\\t=\dfrac{-1+\sqrt3}{\sqrt2}\end{array}\right.$, vì $|t|\le\sqrt2$
Khi đó:
$\Leftrightarrow \left[\begin{array}{l}\sin(x+\dfrac{\pi}{4})=1\\\sin(x+\dfrac{\pi}{4})=\dfrac{-1+\sqrt3}{2}\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\x=\arcsin\dfrac{-1+\sqrt3}{2}-\dfrac{\pi}{4}+k2\pi\\x=\pi-\arcsin\dfrac{-1+\sqrt3}{2}-\dfrac{\pi}{4}+k2\pi\end{array}\right.,k\in\mathbb{Z}$

Phương trình đã cho tương đương với:
      $\cos2x+\sqrt3\sin2x=1$
$\Leftrightarrow \dfrac{1}{2}\cos2x+\dfrac{\sqrt3}{2}\sin2x=\dfrac{1}{2}$
$\Leftrightarrow \sin(2x+\dfrac{\pi}{6})=\sin\dfrac{\pi}{6}$
$\Leftrightarrow \left[\begin{array}{l}2x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{\pi}{3}+k\pi\end{array}\right.,k\in\mathbb{Z}$

Phương trình đã cho tương đương với:
      $\cos x+\sqrt3\sin x=\sqrt3$
$\Leftrightarrow \dfrac{1}{2}\cos x+\dfrac{\sqrt3}{2}\sin x=\dfrac{\sqrt3}{2}$
$\Leftrightarrow \sin(x+\dfrac{\pi}{6})=\sin\dfrac{\pi}{3}$
$\Leftrightarrow \left[\begin{array}{l}x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{array}\right.,k\in\mathbb{Z}$

Phương trình đã cho tương đương với:
      $\sin x+\cos x=\sqrt2 \sin 5x$
$\Leftrightarrow \dfrac{1}{\sqrt2}\sin x+\dfrac{1}{\sqrt2}\cos x=\sin5x$
$\Leftrightarrow \sin(x+\dfrac{\pi}{4})=\sin5x$
$\Leftrightarrow \left[\begin{array}{l}x+\dfrac{\pi}{4}=5x+k2\pi\\x+\dfrac{\pi}{4}=\pi-5x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{16}-k\dfrac{\pi}{2}\\x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}\end{array}\right.,k\in\mathbb{Z}$