•Cosx – Cos2x = Sin3x •Cos7x + Sin8x = Cos3x – Sin2x •Sinx + Sin2x + Sin3x = Cosx + Cos2x + Cos3x

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1. +)

   $\cos x – \cos2x = \sin3x$

   $\cos x – 2\cos²x + 1 = \sin(x + 2x)$

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

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

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

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

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

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

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

   (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:)))

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