+)

$\cos x - \cos2x = \sin3x$

$\cos x - 2\cos²x + 1 = \sin(x + 2x)$

$⇔ -(\cos x - 1)(2\cos x + 1) = \sin x.\cos2x + \cos x.\sin2x$

$⇔ -(\cos x - 1)(2\cos x + 1) = \sin x.(2\cos²x - 1) + 2.\sin x .\cos²x$

$⇔ -(\cos x - 1)(2\cos x + 1) = \sin x.(2\cos²x - 1 + 2cos²x)$

$⇔ -(\cos x - 1)(2\cos x + 1) = \sin x.(4\cos²x - 1)$

$⇔ -(\cos x - 1)(2\cos x + 1) = \sin x(2\cos x - 1)(2\cos x + 1)$

$⇔(2\cos x+1)[\sin x(2\cos x-1)+(\cos x-1)]=0$

$\Leftrightarrow\left[\begin{array}{I}2\cos x+1=0\text{ (1)}\\\sin x(2cos x-1)+(\cos x-1)=0\text{ (2)}\end{array}\right.$

(1) $\Leftrightarrow \cos x=-\dfrac12\Leftrightarrow x=\pm\dfrac{2\pi}3+k2\pi$ $(k\in\mathbb Z)$

(2) $\Leftrightarrow 2\sin x\cos x-\sin x+\cos x-1=0$

$\Leftrightarrow -(\sin x-\cos x)^2-(\sin x-\cos x)=0$

$\Leftrightarrow (\sin x-\cos x)(\sin x-\cos x+1)=0$

$\Leftrightarrow\left[\begin{array}{I}\sin x+\cos x=0\text{ (3)}\\\sin x-\cos x=1\text{ (4)}\end{array}\right.$

(3) $\Leftrightarrow\dfrac1{\sqrt2}\sin\left({x+\dfrac{\pi}4}\right)=0$

$\Leftrightarrow x+\dfrac{\pi}4=k\pi\Leftrightarrow x=-\dfrac{\pi}4+k\pi$ $(k\in\mathbb Z)$

(4) $\Leftrightarrow\dfrac1{\sqrt2}\sin\left({x-\dfrac{\pi}4}\right)=1$

$\Leftrightarrow \sin\left({x-\dfrac{\pi}4}\right)=\sqrt2>1$ (loại)

Vậy phương trình có nghiệm $x=\pm\dfrac{2\pi}3+k2\pi$

và $x=-\dfrac{\pi}4+k\pi$ $(k\in\mathbb Z)$

+)

$\cos7x + \sin8x = \cos3x - \sin2x$
$\Leftrightarrow -2\sin5x\sin2x + 2\sin5x\cos3x=0 $
$\Leftrightarrow -2\sin5x ( \sin2x-\cos3x)=0 $
$\Leftrightarrow \sin 5x = 0$ (1) 
hoặc $\sin2x-\cos3x=0$ (2) 
(1) $\Leftrightarrow 5x = k\pi \Leftrightarrow x= k\dfrac{\pi}5$ $(k\in\mathbb Z)$ 
(2) $\Leftrightarrow \sin 2x = \cos3x$ 
$\Leftrightarrow \sin2x = \sin\left({\dfrac{\pi}2 - 3x}\right)$

$\Leftrightarrow 2x=\dfrac{pi}2 - 3x+k2\pi\Leftrightarrow x=\dfrac{\pi}{10}+k\dfrac{2\pi}{5}$

Hoặc $2x=\pi-\dfrac{\pi}2 + 3x+k2\pi\Leftrightarrow x=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

Vậy phương trình có nghiệm $x= k\dfrac{\pi}5$; $x=\dfrac{\pi}{10}+k\dfrac{2\pi}{5}$; 

$x=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

+

$\sin x + \sin2x + \sin3x = \cos x + \cos2x + \cos3x$
$\Leftrightarrow\sin3x + \sin x + \sin2x = \cos3x + \cos x + \cos2x$
$\Leftrightarrow2\sin2x\cos x + \sin2x = 2\cos2x\cos x + \cos2x$
$\Leftrightarrow\sin 2x(2\cos x + 1) = \cos 2x(2\cos x + 1)$
$\Leftrightarrow(\sin2x - \cos2x)(2\cos x + 1) = 0$
$\Leftrightarrow \sin2x-\cos2x=0\Leftrightarrow \dfrac1{\sqrt2}\sin\left({2x+\dfrac{\pi}4}\right)=0$

$\Leftrightarrow 2x+\dfrac{\pi}4=k\pi\Leftrightarrow x=-\dfrac{\pi}8+k\dfrac{\pi}2$ $(k\in\mathbb Z)$
Hoặc $2\cos x+1=0$

$\Leftrightarrow \cos x=-\dfrac12\Leftrightarrow x= \pm\dfrac{2\pi}3+k2\pi$ $(k\in\mathbb Z)$

Vậy phương trình có nghiệm $ x=-\dfrac{\pi}8+k\dfrac{\pi}2$ và $ x= \pm\dfrac{2\pi}3+k2\pi$ $(k\in\mathbb Z)$

Chúc bạn học tốt:)))