Giải bài 1.15 trang 23 - SBT Đại số và Giải tích lớp 11

\(\begin{aligned} & a)\cos \left( x+3 \right)=\dfrac{1}{3} \\ & \Leftrightarrow \left[ \begin{aligned} & x+3=\arccos \dfrac{1}{3}+k2\pi \\ & x+3=-\arccos \dfrac{1}{3}+k2\pi \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=-3+\arccos \dfrac{1}{3}+k2\pi \\ & x=-3-\arccos \dfrac{1}{3}+k2\pi \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}\)

\(\begin{aligned} & b)\cos \left( 3x-{{45}^{o}} \right)=\dfrac{\sqrt{3}}{2} \\ & \Leftrightarrow \cos \left( 3x-{{45}^{o}} \right)=\cos {{30}^{o}} \\ & \Leftrightarrow \left[ \begin{aligned} & 3x-{{45}^{o}}={{30}^{o}}+k{{.360}^{o}} \\ & 3x-{{45}^{o}}=-{{30}^{o}}+k{{.360}^{o}} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x={{25}^{o}}+k{{.120}^{o}} \\ & x={{5}^{o}}+k{{.120}^{o}} \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}\)

\(\begin{aligned} & c)\cos \left( 2x+\dfrac{\pi }{3} \right)=-\dfrac{1}{2} \\ & \Leftrightarrow \cos \left( 2x+\dfrac{\pi }{3} \right)=\cos \dfrac{2\pi }{3} \\ & \Leftrightarrow \left[ \begin{aligned} & 2x+\dfrac{\pi }{3}=\dfrac{2\pi }{3}+k2\pi \\ & 2x+\dfrac{\pi }{3}=-\dfrac{2\pi }{3}+k2\pi \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\dfrac{\pi }{6}+k\pi \\ & x=-\dfrac{\pi }{2}+k\pi \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}\)

\(\begin{aligned} & d)\left( 2+\cos x \right)\left( 3\cos 2x-1 \right)=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \cos x=-2\,\,(\text{loại}) \\ & \cos 2x=\dfrac{1}{3} \\ \end{aligned} \right. \\ & \Leftrightarrow x=\pm \dfrac{1}{2}\arccos \dfrac{1}{3}+k\pi \,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}\)

Ghi nhớ

\(\cos x=m \\ \Leftrightarrow \left[ \begin{aligned} & x=\alpha+k2\pi \\ & x=-\alpha +k2\pi \\ \end{aligned} \right.\)

với \(\sin \alpha =m\)

\(\cos f(x)=\sin g(x) \\ \Leftrightarrow \left[ \begin{aligned} & f(x)=g(x)+k2\pi \\ & f(x)=-g(x) +k2\pi \\ \end{aligned} \right.\)