Đáp án:

$1)\\ a) \left[\begin{array}{cc} x=\dfrac{\pi}{6}+k 2 \pi & (k \in \mathbb{Z}) \\ x=-\dfrac{\pi}{6}+l 2 \pi & (l \in \mathbb{Z}) \end{array} \right.\\ b) \left[\begin{array}{cc} x= k \pi & (k \in \mathbb{Z})  \\ x=\dfrac{\pi}{2}+ l 2 \pi & (l \in \mathbb{Z})  \end{array} \right.\\ c) \left[\begin{array}{cc}  x= \dfrac{k \pi}{2} & ( k \in \mathbb{Z}) \\ x= \dfrac{3\pi}{8}+ l  \pi & ( l \in \mathbb{Z}) \\ x= -\dfrac{3\pi}{8}+ m \pi & ( m \in \mathbb{Z}) \end{array} \right.\\ d)\left[\begin{array}{cc} x= -\dfrac{\pi}{2} + k 2 \pi & (k \in \mathbb{Z})  \\ x  = \dfrac{\pi}{8} + l  \pi & (l \in \mathbb{Z})\\ x  = -\dfrac{\pi}{8} + m  \pi & (m \in \mathbb{Z}) \end{array} \right. \\ 4)\\ (x+1)^5 =\displaystyle \sum _{k=0}^5 C^k_5 x^kx^{5-k}\\ 5(x+1)^4 =5\displaystyle \sum _{k=0}^4 C^k_4 x^kx^{4-k}$

Giải thích các bước giải:

$1)\\ a)2 \cos x -\sqrt{3}=0\\ \Leftrightarrow 2 \cos x =\sqrt{3}\\ \Leftrightarrow \cos x =\dfrac{\sqrt{3}}{2}\\ \Leftrightarrow \cos x =\cos \left(\dfrac{\pi}{6}\right)\\ \Leftrightarrow \left[\begin{array}{cc} x=\dfrac{\pi}{6}+k 2 \pi & (k \in \mathbb{Z}) \\ x=-\dfrac{\pi}{6}+l 2 \pi & (l \in \mathbb{Z}) \end{array} \right.\\ b)\sin ^2 x - \sin x=0\\ \Leftrightarrow \sin x ( \sin x - 1)=0\\ \Leftrightarrow \left[\begin{array}{l} \sin x =0 \\ \sin x - 1=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin x =0 \\ \sin x = 1 \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x= k \pi & (k \in \mathbb{Z})  \\ x=\dfrac{\pi}{2}+ l 2 \pi & (l \in \mathbb{Z})  \end{array} \right.\\ c)2\sin 2x +\sqrt{2} \sin 4x=0\\ \Leftrightarrow \sqrt{2}\sin 2x + \sin 4x=0\\ \Leftrightarrow \sqrt{2}\sin 2x +2 \sin 2x \cos 2x =0\\ \Leftrightarrow \sin 2x +\sqrt{2} \sin 2x \cos 2x =0\\ \Leftrightarrow \sin 2x \left(1 +\sqrt{2}\cos 2x\right) =0\\ \Leftrightarrow \left[\begin{array}{l} \sin 2x=0 \\ 1 +\sqrt{2}\cos 2x=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin 2x=0 \\ \cos 2x=-\dfrac{\sqrt{2}}{2}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin 2x=0 \\ \cos 2x=\cos \left(\dfrac{3\pi}{4}\right)\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc}  2x= k \pi & ( k \in \mathbb{Z}) \\ \left[\begin{array}{cc}  2x= \dfrac{3\pi}{4}+ l 2 \pi & ( l \in \mathbb{Z}) \\ 2x= -\dfrac{3\pi}{4}+ m 2 \pi & ( m \in \mathbb{Z})  \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc}  x= \dfrac{k \pi}{2} & ( k \in \mathbb{Z}) \\ \left[\begin{array}{cc}  x= \dfrac{3\pi}{8}+ l  \pi & ( l \in \mathbb{Z}) \\ x= -\dfrac{3\pi}{8}+ m \pi & ( m \in \mathbb{Z})  \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc}  x= \dfrac{k \pi}{2} & ( k \in \mathbb{Z}) \\ x= \dfrac{3\pi}{8}+ l  \pi & ( l \in \mathbb{Z}) \\ x= -\dfrac{3\pi}{8}+ m \pi & ( m \in \mathbb{Z}) \end{array} \right.\\ d)(\sin x+1)(2 \cos 2x - \sqrt{2} )=0\\ \Leftrightarrow \left[\begin{array}{l} \sin x+1 =0 \\ 2 \cos 2x - \sqrt{2}  =0 \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sin x =-1 \\ \cos 2x  =\dfrac{\sqrt{2}}{2} \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x= -\dfrac{\pi}{2} + k 2 \pi & (k \in \mathbb{Z}) \\ \cos 2x  =\cos \left(\dfrac{\pi}{4}\right)\end{array} \right. \\ \Leftrightarrow \left[\begin{array}{cc} x= -\dfrac{\pi}{2} + k 2 \pi & (k \in \mathbb{Z})  \\ \left[\begin{array}{cc} 2x  = \dfrac{\pi}{4} + l 2 \pi & (l \in \mathbb{Z})\\ 2x  = -\dfrac{\pi}{4} + m 2 \pi & (m \in \mathbb{Z})\end{array} \right.  \end{array} \right. \\ \Leftrightarrow \left[\begin{array}{cc} x= -\dfrac{\pi}{2} + k 2 \pi & (k \in \mathbb{Z})  \\ \left[\begin{array}{cc} x  = \dfrac{\pi}{8} + l  \pi & (l \in \mathbb{Z})\\ x  = -\dfrac{\pi}{8} + m  \pi & (m \in \mathbb{Z})\end{array} \right.  \end{array} \right. \\ \Leftrightarrow \left[\begin{array}{cc} x= -\dfrac{\pi}{2} + k 2 \pi & (k \in \mathbb{Z})  \\ x  = \dfrac{\pi}{8} + l  \pi & (l \in \mathbb{Z})\\ x  = -\dfrac{\pi}{8} + m  \pi & (m \in \mathbb{Z}) \end{array} \right. \\ 4)\\ (x+1)^5\\ =\displaystyle \sum _{k=0}^5 C^k_5 x^kx^{5-k}\\ =C^0_5 x^0x^{5}+\cdots + C^5_5 x^5x^{0}\\ 5(x+1)^4\\ =5\displaystyle \sum _{k=0}^4 C^k_4 x^kx^{4-k}\\ =5\left(C^0_4 x^0x^{4}+\cdots + C^4_4 x^4x^{0}\right)$