Điều kiện: $\sin 2x\ne-1 \Leftrightarrow x\ne\dfrac{-\pi}{4}+k\pi,k\in\mathbb{Z}$
Phương trình đã cho tương đương với:
      $2\sin x\cos x+3\sqrt2\cos x-2\cos^2x-1=1+\sin2x$
$\Leftrightarrow \sin2x+3\sqrt2\cos x-2\cos^2x-2=\sin2x$
$\Leftrightarrow 2\cos^2x-3\sqrt2\cos x+2=0$
$\Leftrightarrow \cos x=\dfrac{1}{\sqrt2}$, vì $|\cos x|\le1$
$\Leftrightarrow x=\pm\dfrac{\pi}{4}+k2\pi,k\in\mathbb{Z}$
Kết hợp đk, ta có: $x=\dfrac{\pi}{4}+k2\pi,k\in\mathbb{Z}$

Điều kiện: $\cos x\ne0 \Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}$
Phương trình đã cho tương đương với:
      $\tan^3x-1+\tan^2x+1-3\tan x=3$
$\Leftrightarrow \tan^3x+\tan^2x-3\tan x-3=0$
$\Leftrightarrow \left[\begin{array}{l}\tan x=-1\\\tan x=\sqrt3\\\tan x=-\sqrt3\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{-\pi}{4}+k\pi\\x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{-\pi}{3}+k\pi\end{array}\right.,k\in\mathbb{Z}$, thỏa mãn.

a.
Ta có:
$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$
Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{2(\sin A+\sin B+\sin C)}$
Mà ta có:
$\sin A+\sin B+\sin C= 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}+2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$
                                          $=2\cos\dfrac{A}{2}(\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2})$
                                          $=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$
$\Rightarrow \dfrac{r}{4R}=\dfrac{8\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}{8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}=\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$

Theo định lý hàm số sin, ta có:
      $16R^4(c^4+2a^4+2b^4)=32R^4c^2(a^2+b^2)$
$\Leftrightarrow c^4+2a^4+2b^4=2c^2(a^2+b^2)$
$\Leftrightarrow (c^2-a^2-b^2)^2+(a^2-b^2)^2=0$
$\Leftrightarrow \left\{\begin{array}{l}c^2=a^2+b^2\\a^2=b^2\end{array}\right.\Leftrightarrow \Delta ABC$ vuông cân tại $C$.

b.
Theo phần (a) ta có: $a+b=3c \Leftrightarrow p=2c          (1)$
Mặt khác:
$S=pr=\dfrac{abc}{4R}\Rightarrow pabc\dfrac{r}{4R}=S^2=p(p-a)(p-b)(p-c)$
Thay $(1)$ vào biểu thức trên ta được:
$ab\dfrac{r}{4R}=p^2-(a+b)p+ab\Rightarrow (1-\dfrac{r}{4R})ab=2c^2     (2)$
Lại có:
$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$
Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{2(\sin A+\sin B+\sin C)}$
Mà ta có:
$\sin A+\sin B+\sin C= 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}+2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$
                                          $=2\cos\dfrac{A}{2}(\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2})$
                                          $=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$
$\Rightarrow \dfrac{r}{4R}=\dfrac{8\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}{8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}=\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$
$\Rightarrow \dfrac{r}{4R}=\dfrac{1}{10}$
Thay vào $(2)$ ta được: $ab=\dfrac{20}{9}c^2$
$\Rightarrow ab=20(a+b)^2$
$\Leftrightarrow 20a^2-41ab+20b^2=0$
$\Leftrightarrow \left\{\begin{array}{l}4a=5b\\5a=4b\end{array}\right.$
Cả 2 trường hợp đều dẫn tới $\Delta ABc$ vuông. (theo Pytago)

a.
Ta có:
$S=p(p-a)\tan\dfrac{A}{2}=p(p-b)\tan\dfrac{B}{2}=\sqrt{p(p-a)(p-b)(p-c)}$
$\Rightarrow \tan\dfrac{A}{2}\tan\dfrac{B}{2}=\dfrac{S^2}{p^2(p-a)(p-b)}=\dfrac{p-c}{p}=\dfrac{a+b-c}{a+b+c}$
Mà: $\tan\dfrac{A}{2}\tan\dfrac{B}{2}=\dfrac{1}{2}\Leftrightarrow \dfrac{a+b-c}{a+b+c}=\dfrac{1}{2}\Leftrightarrow a+b=3c$

Phương trình đã cho tương đương với:
      $(2\sin x-1)(2\sin2x+1)=4\sin^2x-1$
$\Leftrightarrow (2\sin x-1)(2\sin2x+1)=(2\sin x-1)(2\sin x+1)$
$\Leftrightarrow 2(2\sin x-1)(\sin2x-\sin x)=0$
$\Leftrightarrow \left[\begin{array}{l}\sin x=\dfrac{1}{2}\\\sin2x=\sin x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=k2\pi\\x=\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\end{array}\right.,k\in\mathbb{Z}$

2.
Phương trình đã cho tương đương với:
      $\sin9x-\sqrt3\cos9x=\sin7x-\sqrt3\cos7x$
$\Leftrightarrow \dfrac{1}{2}\sin9x-\dfrac{\sqrt3}{2}\cos9x=\dfrac{1}{2}\sin7x-\dfrac{\sqrt3}{2}\cos7x$
$\Leftrightarrow \sin(9x-\dfrac{\pi}{3})=\sin(7x-\dfrac{\pi}{3})$
$\Leftrightarrow \left[\begin{array}{l}9x-\dfrac{\pi}{3}=7x-\dfrac{\pi}{3}+k2\pi\\9x-\dfrac{\pi}{3}=\dfrac{4\pi}{3}-7x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{5\pi}{48}+k\dfrac{\pi}{8}\end{array}\right.,k\in\mathbb{Z}$

1.
Phương trình đã cho tương đương với:
      $\cos x-\sqrt3\sin x=2\cos3x$
$\Leftrightarrow \dfrac{1}{2}\cos x-\dfrac{\sqrt3}{2}\sin x=\cos3x$
$\Leftrightarrow \cos(x+\dfrac{\pi}{3})=\cos3x$
$\Leftrightarrow \left[\begin{array}{l}x+\dfrac{\pi}{3}=3x+k2\pi\\x+\dfrac{\pi}{3}=-3x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{6}-k\pi\\x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2}\end{array}\right.,k\in\mathbb{Z}$